Class 11 Chemistry Chapter 2 – NCERT Solutions (Structure of the Atom)

NCERT Solutions for Class 11 Chemistry Chapter 2 gives you step-by-step answers for all questions from Structure of the Atom. This page explains topics like Bohr Model, Quantum Numbers, Electronic Configuration, de Broglie, and Heisenberg in simple language. It is very helpful for CBSE/NCERT preparation, school tests, and board exams.

Keywords: Class 11 Chemistry Chapter 2 NCERT Solutions, Structure of the Atom, परमाणु की संरचना, NCERT solutions, CBSE Class 11 Chemistry, Hindi Medium

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Class 11 Chemistry Chapter 2 NCERT Solutions – Structure of Atom (परमाणु की संरचना)

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Q.1: (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.

Answer

✅ (i) Number of electrons in 1 g of electrons ≈ 1.10 × 10²⁷ electrons
✅ (ii) Mass of 1 mole of electrons ≈ 5.49 × 10⁻⁴ g (or 5.49 × 10⁻⁷ kg)
✅ Total charge of 1 mole of electrons ≈ −9.65 × 10⁴ C (approximately −96485 C)

Explanation – Step by Step

Given:
👉 Mass of 1 electron (m_e) = 9.11 × 10⁻³¹ kg = 9.11 × 10⁻²⁸ g

(i) Number of electrons in 1 g
Formula: Number of electrons = (total mass) / (mass of 1 electron)

N = 1 g / (9.11 × 10⁻²⁸ g) ≈ 1.10 × 10²⁷ electrons

(ii) Mass and charge of 1 mole of electrons

(A) Mass of 1 mole of electrons
Given:
👉 N_A (Avogadro’s number) = 6.022 × 10²³ mol⁻¹
👉 Mass of 1 electron = 9.11 × 10⁻³¹ kg

Mass of 1 mole of electrons = (mass of 1 electron) × (N_A)
= 9.11 × 10⁻³¹ kg × 6.022 × 10²³
≈ 5.49 × 10⁻⁷ kg
= 5.49 × 10⁻⁴ g

(B) Charge of 1 mole of electrons
Given:
👉 Charge of 1 electron (e) = 1.602 × 10⁻¹⁹ C
👉 1 mole of electrons = N_A electrons

Total charge = −e × N_A
= −(1.602 × 10⁻¹⁹) × (6.022 × 10²³)
≈ −9.65 × 10⁴ C
≈ −96485 C

Did You Know?

👉 The mass of an electron is extremely small, so 1 gram contains a very large number of electrons (around 10²⁷).
👉 The charge of 1 mole of electrons is equal to 1 Faraday. Its value is approximately 96485 C mol⁻¹, which is very useful in electrolysis problems.

Q.2: (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C. (Assume that mass of a neutron = 1.675 × 10^–27 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP. Will the answer change if the temperature and pressure are changed?

Answer

✅ (i) Electrons in 1 mole of CH₄ = 6.022 × 10²⁴

✅ (ii) In 7 mg of ¹⁴C:
👉 (a) Total neutrons = 2.4088 × 10²¹
👉 (b) Total mass of neutrons = 4.03474 × 10⁻⁶ kg (≈ 4.03 mg)

✅ (iii) In 34 mg of NH₃:
👉 (a) Total protons = 1.2044 × 10²²
👉 (b) Total mass of protons ≈ 2.015 × 10⁻⁵ kg (≈ 20.15 mg)
👉 On changing temperature/pressure: Answer will not change (because mass is given)

Explanation – Step by Step

(i) Number of electrons in 1 mole of CH₄
👉 Electrons in C = 6
👉 Electrons in H = 1, and there are 4 H atoms ⇒ 4 × 1 = 4
⇒ Total electrons in 1 molecule of CH₄ = 6 + 4 = 10

Now, molecules in 1 mole of CH₄ = N_A = 6.022 × 10²³
⇒ Total electrons = 10 × N_A
= 10 × 6.022 × 10²³ = 6.022 × 10²⁴
________________________________________

(ii) Neutrons in 7 mg of ¹⁴C (number and mass)

Step 1: Neutrons in one atom of ¹⁴C
Atomic number (Z) of ¹⁴C = 6
Neutrons = A − Z = 14 − 6 = 8

Step 2: Number of atoms in 7 mg of ¹⁴C
Mass of ¹⁴C = 7 mg = 0.007 g
Molar mass of ¹⁴C = 14 g mol⁻¹
Moles = mass ÷ molar mass
Moles = 0.007 / 14 = 5.0 × 10⁻⁴ mol

Atoms = (5.0 × 10⁻⁴) × (6.022 × 10²³)
= 3.011 × 10²⁰ atoms

Step 3: Total neutrons
Neutrons per atom = 8
Total neutrons = 8 × 3.011 × 10²⁰
= 2.4088 × 10²¹

Step 4: Total mass of neutrons
Mass of 1 neutron = 1.675 × 10⁻²⁷ kg
Total mass = (2.4088 × 10²¹) × (1.675 × 10⁻²⁷)
= 4.03474 × 10⁻⁶ kg
= 4.03 × 10⁻³ g ≈ 4.03 mg
________________________________________

(iii) Protons in 34 mg of NH₃ (number and mass)

Step 1: Protons in 1 molecule of NH₃
Atomic number of N = 7 ⇒ 7 protons
Atomic number of H = 1 and there are 3 H atoms ⇒ 3 protons
⇒ Total protons per molecule = 7 + 3 = 10

Step 2: Number of molecules in 34 mg NH₃
Mass of NH₃ = 34 mg = 0.034 g
Molar mass of NH₃ = 17 g mol⁻¹
Moles = mass ÷ molar mass
Moles = 0.034 / 17 = 2.0 × 10⁻³ mol

Molecules = (2.0 × 10⁻³) × (6.022 × 10²³)
= 1.2044 × 10²¹ molecules

Step 3: Total protons
Total protons = 10 × 1.2044 × 10²¹
= 1.2044 × 10²²

Step 4: Total mass of protons
(Standard value) mass of 1 proton ≈ 1.673 × 10⁻²⁷ kg
Total mass = (1.2044 × 10²²) × (1.673 × 10⁻²⁷)
≈ 2.015 × 10⁻⁵ kg
= 2.015 × 10⁻² g ≈ 20.15 mg

Will the answer change with temperature and pressure?
Since the sample is given by mass (34 mg), moles and number of particles remain the same.
✅ Therefore, total number and mass of protons will not change.

Did You Know?

👉 The quickest way to find total electrons/protons/neutrons is:
1) First find particles in 1 molecule/1 atom
2) Then calculate moles of the sample
3) Multiply moles by Avogadro’s number (6.022 × 10²³) to get total particles

Q.3: How many neutrons and protons are there in the following nuclei?
⁸⁸Sr₃₈, ⁵⁶Fe₂₆, ²⁴Mg₁₂, ¹⁶O₈, ¹³C₆

Answer

✅ In ⁸⁸Sr₃₈: Protons = 38, Neutrons = 50
✅ In ⁵⁶Fe₂₆: Protons = 26, Neutrons = 30
✅ In ²⁴Mg₁₂: Protons = 12, Neutrons = 12
✅ In ¹⁶O₈: Protons = 8, Neutrons = 8
✅ In ¹³C₆: Protons = 6, Neutrons = 7

Explanation – Step by Step

Rule:
👉 Atomic number (Z) = Number of protons
👉 Mass number (A) = Protons + Neutrons
👉 Therefore, Number of neutrons = A − Z

Nucleus	Mass Number (A)	Atomic Number (Z)	Protons (Z)	Neutrons (A − Z)	Calculation
⁸⁸Sr₃₈	88	38	38	50	88 − 38 = 50
⁵⁶Fe₂₆	56	26	26	30	56 − 26 = 30
²⁴Mg₁₂	24	12	12	12	24 − 12 = 12
¹⁶O₈	16	8	8	8	16 − 8 = 8
¹³C₆	13	6	6	7	13 − 6 = 7

Did You Know?

👉 For any nucleus, remember in one line: Z = protons, and A − Z = neutrons.
👉 In isotopes (like ¹²C and ¹³C), protons stay the same and only neutrons change.

Q.4: Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A):
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9

Answer

✅ (i) ³⁵Cl₁₇
✅ (ii) ²³³U₉₂
✅ (iii) ⁹Be₄

Explanation – Step by Step

Rule for writing complete atomic symbol:
Complete atomic symbol = ^AX_Z

Where:
👉 A = Mass number
👉 Z = Atomic number
👉 X = Element symbol

(i) Z = 17, A = 35
Element with Z = 17 is Cl (Chlorine)
So, complete symbol = ³⁵Cl₁₇

(ii) Z = 92, A = 233
Element with Z = 92 is U (Uranium)
So, complete symbol = ²³³U₉₂

(iii) Z = 4, A = 9
Element with Z = 4 is Be (Beryllium)
So, complete symbol = ⁹Be₄

Did You Know?

👉 Atomic number (Z) identifies the element because Z = number of protons.
👉 Mass number (A) is equal to protons + neutrons.

Q.5: Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber (ν̄) of the yellow light.

Answer

✅ Frequency (ν) = 5.17 × 10¹⁴ s⁻¹
✅ Wavenumber (ν̄) = 1.72 × 10⁶ m⁻¹

Explanation – Step by Step

Given:
λ = 580 nm = 580 × 10⁻⁹ m
Speed of light (c) = 3.0 × 10⁸ m s⁻¹

(1) Frequency (ν)
Formula: ν = c / λ

ν = (3.0 × 10⁸) / (580 × 10⁻⁹)
ν = (3.0/580) × 10¹⁷
ν ≈ 0.00517 × 10¹⁷
ν = 5.17 × 10¹⁴ s⁻¹

✅ Therefore, frequency (ν) = 5.17 × 10¹⁴ s⁻¹

(2) Wavenumber (ν̄)
Formula for wavenumber: ν̄ = 1 / λ

ν̄ = 1 / (580 × 10⁻⁹) m⁻¹
ν̄ = (1/580) × 10⁹ m⁻¹
ν̄ ≈ 0.001724 × 10⁹ m⁻¹
ν̄ = 1.72 × 10⁶ m⁻¹

✅ Therefore, wavenumber (ν̄) = 1.72 × 10⁶ m⁻¹

Did You Know?

👉 As wavelength increases, frequency decreases, because ν = c/λ.
👉 Wavenumber (ν̄) is also inversely proportional to λ, so smaller λ gives larger ν̄.

Q.6: Find the energy of each of the photons which:
(i) correspond to light of frequency 3 × 10¹⁵ Hz.
(ii) have wavelength of 0.50 Å.

Answer

✅ (i) Energy = 1.99 × 10⁻¹⁸ J (approx.)
✅ (ii) Energy = 3.98 × 10⁻¹⁵ J (approx.)

Explanation – Step by Step

Constants:
h = 6.626 × 10⁻³⁴ J s
c = 3.0 × 10⁸ m s⁻¹

(i) When frequency (ν) is given
Formula: E = hν
ν = 3 × 10¹⁵ s⁻¹

E = (6.626 × 10⁻³⁴) × (3 × 10¹⁵)
E = 19.878 × 10⁻¹⁹ J
E = 1.9878 × 10⁻¹⁸ J
E ≈ 1.99 × 10⁻¹⁸ J

✅ Therefore, energy = 1.99 × 10⁻¹⁸ J (approx.)

(ii) When wavelength (λ) is given
Formula: E = hc/λ

First convert λ into meter (m):
1 Å = 10⁻¹⁰ m
λ = 0.50 Å = 0.50 × 10⁻¹⁰ m = 5.0 × 10⁻¹¹ m

Now calculate energy:
E = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / (5.0 × 10⁻¹¹)
E = (19.878 × 10⁻²⁶) / (5.0 × 10⁻¹¹)
E = 3.9756 × 10⁻¹⁵ J
E ≈ 3.98 × 10⁻¹⁵ J

✅ Therefore, energy = 3.98 × 10⁻¹⁵ J (approx.)

Did You Know?

👉 Higher the frequency, higher is the photon energy.
👉 Radiation with very small wavelength (like Å range, e.g., X-rays) has very high energy.

Q.7: Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 × 10⁻¹⁰ s.

Answer

✅ Frequency (ν) = 5.0 × 10⁹ s⁻¹
✅ Wavelength (λ) = 6.0 × 10⁻² m
✅ Wavenumber (ν̄) = 16.67 m⁻¹

Explanation – Step by Step

Given:
Time period (T) = 2.0 × 10⁻¹⁰ s
Speed of light (c) = 3.0 × 10⁸ m s⁻¹

(1) Frequency (ν)
Formula: ν = 1/T

ν = 1 / (2.0 × 10⁻¹⁰)
ν = (1/2.0) × 10¹⁰
ν = 0.5 × 10¹⁰
ν = 5.0 × 10⁹ s⁻¹
________________________________________

(2) Wavelength (λ)
Formula: λ = c/ν

λ = (3.0 × 10⁸) / (5.0 × 10⁹)
λ = (3.0/5.0) × 10⁻¹
λ = 0.6 × 10⁻¹
λ = 6.0 × 10⁻² m
________________________________________

(3) Wavenumber (ν̄)
Formula: ν̄ = 1/λ

ν̄ = 1 / (6.0 × 10⁻²)
ν̄ = (1/6.0) × 10²
ν̄ = 0.1667 × 10²
ν̄ = 16.67 m⁻¹

Did You Know?

👉 Time period (T) and frequency (ν) are reciprocals of each other: ν = 1/T.
👉 For a light wave, the relation is always: c = νλ.

Q.8: What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Answer

✅ Number of photons required to provide 1 J energy ≈ 2.01 × 10¹⁶ photons

Explanation – Step by Step

Given:
👉 Wavelength (λ) = 4000 pm
👉 Speed of light (c) = 3.0 × 10⁸ m s⁻¹
👉 Planck’s constant (h) = 6.626 × 10⁻³⁴ J s
👉 Total energy = 1 J

Step 1: Convert λ into meter (m)
1 pm = 10⁻¹² m
λ = 4000 × 10⁻¹² m = 4.0 × 10⁻⁹ m
________________________________________

Step 2: Find frequency (ν)
Formula: ν = c/λ

ν = (3.0 × 10⁸) / (4.0 × 10⁻⁹)
ν = (3.0/4.0) × 10¹⁷
ν = 0.75 × 10¹⁷
ν = 7.5 × 10¹⁶ s⁻¹
________________________________________

Step 3: Find energy of one photon (E)
Formula: E = hν

E = (6.626 × 10⁻³⁴) × (7.5 × 10¹⁶) J
E = 49.695 × 10⁻¹⁸ J
E = 4.97 × 10⁻¹⁷ J (per photon)
________________________________________

Step 4: Number of photons (N) for 1 J energy
Formula: N = (Total energy) / (Energy of one photon)

N = 1 / (4.97 × 10⁻¹⁷)
N ≈ 2.01 × 10¹⁶

✅ Therefore, number of photons ≈ 2.01 × 10¹⁶

Did You Know?

👉 Smaller the wavelength (λ), higher is the energy of one photon.
👉 That is why short-wavelength radiation (like UV and X-rays) can deliver high energy even with fewer photons.

Q.9: A photon of wavelength 4 × 10⁻⁷ m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV = 1.6020 × 10⁻¹⁹ J).

Answer

✅ (i) Energy of photon = 3.10 eV
✅ (ii) Kinetic energy of electron = 0.97 eV (or 1.56 × 10⁻¹⁹ J)
✅ (iii) Velocity of photoelectron = 5.85 × 10⁵ m s⁻¹

Explanation – Step by Step

Given:
👉 λ = 4 × 10⁻⁷ m, c = 3.0 × 10⁸ m s⁻¹
👉 h = 6.626 × 10⁻³⁴ J s
👉 Work function (W) = 2.13 eV
👉 1 eV = 1.6020 × 10⁻¹⁹ J
👉 Mass of electron (m) = 9.109 × 10⁻³¹ kg

Step 1: Calculate frequency (ν)
Formula: ν = c/λ

ν = (3.0 × 10⁸) / (4 × 10⁻⁷)
ν = (3.0/4) × 10¹⁵
ν = 7.5 × 10¹⁴ s⁻¹

Step 2: Energy of photon (in J), then convert to eV
Formula: E = hν

E = (6.626 × 10⁻³⁴) × (7.5 × 10¹⁴)
E = 49.695 × 10⁻²⁰ J
E = 4.97 × 10⁻¹⁹ J

Now convert energy to eV:
E(eV) = E(J) / (1.6020 × 10⁻¹⁹)
E(eV) = (4.97 × 10⁻¹⁹) / (1.6020 × 10⁻¹⁹)
E(eV) = 4.97/1.6020 = 3.10 eV

✅ Therefore, energy of photon = 3.10 eV

Step 3: Kinetic energy (KE) of emitted electron
Einstein’s equation:
Photon energy (E) = Work function (W) + Kinetic energy (KE)

KE = E − W
KE = 3.10 eV − 2.13 eV = 0.97 eV

Convert KE into J:
KE(J) = 0.97 × (1.6020 × 10⁻¹⁹)
KE(J) = 1.56 × 10⁻¹⁹ J

✅ Therefore, KE = 0.97 eV (or 1.56 × 10⁻¹⁹ J)

Step 4: Velocity (v) of photoelectron
Formula: KE = (1/2)mv²
So, v = √(2KE/m)

v = √( (2 × 1.56 × 10⁻¹⁹) / (9.109 × 10⁻³¹) )
v = √(3.12/9.109 × 10¹²)
v = √(0.3426 × 10¹²)
v = √(3.426 × 10¹¹)
v = 5.85 × 10⁵ m s⁻¹

✅ Therefore, velocity of photoelectron = 5.85 × 10⁵ m s⁻¹

Did You Know?

👉 If photon energy (E) is less than work function (W), no electron is emitted (photoelectric effect will not occur).
👉 Higher E gives higher kinetic energy and hence higher electron speed.

Q.10: Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol⁻¹.

Answer

✅ Ionisation energy of sodium = 494.6 kJ mol⁻¹ (approx.)

Explanation – Step by Step

Given:
👉 λ = 242 nm = 242 × 10⁻⁹ m
👉 h = 6.626 × 10⁻³⁴ J s
👉 c = 3.0 × 10⁸ m s⁻¹
👉 N_A = 6.022 × 10²³ mol⁻¹
________________________________________

Step 1: Energy for 1 photon (or 1 atom)
Formula: E = hc/λ

E = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / (242 × 10⁻⁹)
E = (19.878 × 10⁻²⁶) / (242 × 10⁻⁹)
E = (19.878/242) × 10⁻¹⁷
E = 0.08214 × 10⁻¹⁷
E = 8.214 × 10⁻¹⁹ J (per atom)
________________________________________

Step 2: Ionisation energy for 1 mole
Number of atoms in 1 mole = N_A

Ionisation energy (J mol⁻¹)
= (8.214 × 10⁻¹⁹) × (6.022 × 10²³)
= (8.214 × 6.022) × 10⁴
= 49.46 × 10⁴ J mol⁻¹
= 4.946 × 10⁵ J mol⁻¹

Now convert to kJ mol⁻¹:
4.946 × 10⁵ J mol⁻¹ = 494.6 kJ mol⁻¹

✅ Therefore, ionisation energy of sodium = 494.6 kJ mol⁻¹ (approx.)

Did You Know?

👉 To find ionisation energy, remember 2 steps:
1) First find energy of one photon/one atom in joules.
2) Then multiply by Avogadro’s number to get energy for one mole.

Q.11: A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.

Answer

✅ Number of quanta (photons) emitted per second = 7.17 × 10¹⁹ s⁻¹ (approximately)

Explanation – Step by Step

Given:
👉 Power (P) = 25 W = 25 J s⁻¹
👉 Wavelength (λ) = 0.57 µm = 0.57 × 10⁻⁶ m
👉 h = 6.626 × 10⁻³⁴ J s
👉 c = 3.0 × 10⁸ m s⁻¹
Step 1: Energy of one photon (E)
Formula: E = hc/λ

E = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / (0.57 × 10⁻⁶)
E = (19.878 × 10⁻²⁶) / (0.57 × 10⁻⁶)
E = (19.878/0.57) × 10⁻²⁰
E = 34.87 × 10⁻²⁰ J
E = 3.49 × 10⁻¹⁹ J (per photon)

Step 2: Number of photons emitted per second (N)
Power (P) = energy emitted per second
Therefore, N = P/E

N = 25 / (3.49 × 10⁻¹⁹)
N = (25/3.49) × 10¹⁹
N ≈ 7.17 × 10¹⁹ s⁻¹

✅ Therefore, photons emitted per second ≈ 7.17 × 10¹⁹ s⁻¹

Did You Know?

👉 1 watt = 1 joule/second. So 25 W means 25 J energy is emitted every second.
👉 Lower energy per photon means more photons are emitted per second for the same power.

Q.12: Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate the threshold frequency (ν₀) and work function (W₀) of the metal.

Answer

✅ Threshold frequency (ν₀) = 4.41 × 10¹⁴ s⁻¹
✅ Work function (W₀) = 2.92 × 10⁻¹⁹ J ≈ 1.82 eV

Explanation – Step by Step

Given:
λ₀ = 6800 Å
1 Å = 10⁻¹⁰ m
λ₀ = 6800 × 10⁻¹⁰ m = 6.8 × 10⁻⁷ m
c = 3.0 × 10⁸ m s⁻¹
h = 6.626 × 10⁻³⁴ J s
1 eV = 1.602 × 10⁻¹⁹ J

👉 Here, “zero velocity” means threshold condition, so this λ₀ is the threshold wavelength.
________________________________________

(1) Threshold frequency (ν₀)
Formula: ν₀ = c / λ₀

ν₀ = (3.0 × 10⁸) / (6.8 × 10⁻⁷)
ν₀ = (3.0/6.8) × 10¹⁵
ν₀ = 0.441 × 10¹⁵
ν₀ = 4.41 × 10¹⁴ s⁻¹
________________________________________

(2) Work function (W₀)
At threshold: W₀ = hν₀

W₀ = (6.626 × 10⁻³⁴) × (4.41 × 10¹⁴) J
W₀ = (6.626 × 4.41) × 10⁻²⁰ J
W₀ = 29.22 × 10⁻²⁰ J
W₀ = 2.92 × 10⁻¹⁹ J

Now convert into eV:
W₀(eV) = W₀(J) / (1.602 × 10⁻¹⁹)
W₀(eV) = (2.92 × 10⁻¹⁹) / (1.602 × 10⁻¹⁹)
W₀(eV) = 2.92/1.602 = 1.82 eV

✅ Therefore, W₀ = 2.92 × 10⁻¹⁹ J ≈ 1.82 eV

Did You Know?

👉 At threshold wavelength (λ₀), an electron just starts to come out, so its kinetic energy is zero.
👉 If λ < λ₀ (shorter wavelength), emitted electrons get higher kinetic energy because photon energy is higher.

Q.13: जब हाइड्रोजन परमाणु के n = 4 ऊर्जा स्तर से n = 2 ऊर्जा स्तर में इलेक्ट्रॉन जाता है, तो किस तरंगदैर्घ्य का प्रकाश उत्सर्जित होगा?

उत्तर

✅ उत्सर्जित प्रकाश की तरंगदैर्घ्य (λ) = 4.87 × 10⁻⁷ m = 487 nm (लगभग)

व्याख्या – Step by Step

हाइड्रोजन के nवें कक्ष की ऊर्जा:
E_n = −(2.178 × 10⁻¹⁸) / n² J atom⁻¹

दिया है:
👉 n₁ = 4, n₂ = 2
👉 h = 6.626 × 10⁻³⁴ J s
👉 c = 3.0 × 10⁸ m s⁻¹
________________________________________

Step 1: E₂ और E₄ निकालें

E₂ = −(2.178 × 10⁻¹⁸) / (2²)
E₂ = −(2.178 × 10⁻¹⁸) / 4
E₂ = −5.445 × 10⁻¹⁹ J atom⁻¹

E₄ = −(2.178 × 10⁻¹⁸) / (4²)
E₄ = −(2.178 × 10⁻¹⁸) / 16
E₄ = −1.361 × 10⁻¹⁹ J atom⁻¹
________________________________________

Step 2: ऊर्जा अंतर (ΔE)
उत्सर्जित ऊर्जा = |E₂ − E₄|

ΔE = |(−5.445 × 10⁻¹⁹) − (−1.361 × 10⁻¹⁹)|
ΔE = |−4.084 × 10⁻¹⁹|
ΔE = 4.08 × 10⁻¹⁹ J atom⁻¹

(यह उसी के बराबर है: 2.178 × 10⁻¹⁸ × (1/4 − 1/16) = 2.178 × 10⁻¹⁸ × (3/16))
________________________________________

Step 3: तरंगदैर्घ्य (λ)
ΔE = hc/λ ⇒ λ = hc/ΔE

λ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / (4.08 × 10⁻¹⁹)
λ = (19.878 × 10⁻²⁶) / (4.08 × 10⁻¹⁹)
λ = (19.878/4.08) × 10⁻⁷
λ = 4.87 × 10⁻⁷ m

nm में:
4.87 × 10⁻⁷ m = 4.87 × 10² nm = 487 nm

✅ इसलिए उत्सर्जित प्रकाश की तरंगदैर्घ्य = 4.87 × 10⁻⁷ m = 487 nm

क्या आप जानते हैं?

👉 n = 4 से n = 2 में गिरने पर जो रेखा आती है, वह Balmer series में आती है।
👉 Balmer series की रेखाएँ मुख्यतः दृश्य (visible) क्षेत्र में होती हैं, इसलिए 487 nm वाली रेखा आँखों से देखी जा सकती है।

Q.14: How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (from n = 1 orbit).

Answer

👉 Energy required to ionise H atom from n = 5 is:
8.72 × 10^-20 J atom^-1
(or 5.25 × 10⁴ J mol^-1 = 52.5 kJ mol^-1)

👉 Ionization enthalpy from n = 1 is:
2.18 × 10^-18 J atom^-1
(or 13.12 × 10⁵ J mol^-1 = 1312 kJ mol^-1)

👉 Comparison:
Ionization from n = 1 is 25 times larger than from n = 5. ✅

Explanation (Step-by-Step)

📌 For hydrogen atom, energy in any orbit is:
E_n = -2.18 × 10^-18/n² J atom^-1

Step 1) Energy of electron in n = 5 orbit
👉 E₅ = -2.18 × 10^-18/25
👉 E₅ = -8.72 × 10^-20 J atom^-1

Step 2) Ionization energy from n = 5
📌 Ionization means taking electron to n = ∞, where E_∞ = 0.
So,
IE₅ = E_∞ - E₅
IE₅ = 0 - ( -8.72 × 10^-20 )
✅ IE₅ = 8.72 × 10^-20 J atom^-1

Step 3) Compare with ionization from n = 1
For n = 1:
E₁ = -2.18 × 10^-18 J atom^-1
So ionization energy from ground state = 2.18 × 10^-18 J atom^-1

Now ratio:
IE₁/IE₅ = (2.18 × 10^-18)/(8.72 × 10^-20) = 25

✅ Ground-state ionization enthalpy is 25 times the n = 5 value.

Did You Know?

💡 Higher orbit (larger n) means electron is farther from nucleus, so it is less tightly held.
💡 That is why very little energy is needed to remove electron from n = 5 compared to n = 1.
💡 In excited states, hydrogen ionizes much more easily. ⚡

Q.15: What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Answer

👉 The maximum number of emission lines is 15 ✅

Explanation (Step-by-Step)

For an electron falling from level n to lower levels, maximum possible emission lines are:
N=(n(n-1))/2

Now put n = 6:
N=6(6-1)/2
=(6×5)/2
=15

👉 So total possible distinct transitions = 15 lines .

________________________________________

All possible transitions from n = 6

📌 From 6th level:
   6 → 5
   6 → 4
   6 → 3
   6 → 2
   6 → 1 (5 lines)

📌 From 5th level:
   5 → 4
   5 → 3
   5 → 2
   5 → 1 (4 lines)

📌 From 4th level:
   4 → 3
   4 → 2
   4 → 1 (3 lines)

📌 From 3rd level:
   3 → 2
   3 → 1 (2 lines)

📌 From 2nd level:
   2 → 1 (1 line)

👉 Total = 5 + 4 + 3 + 2 + 1 = 15

Did You Know?

💡 Not all lines appear in the same spectral region.
💡 Transitions ending at:
   n = 1 belong to Lyman series (UV)
   n = 2 belong to Balmer series (Visible)
   n = 3 belong to Paschen series (IR)
So one excited state can generate lines in different series too. 🌈

Q.16: (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10^–18 J atom^–1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Answer

👉 (i) Energy of 5th orbit, E₅ = –8.72 × 10^–20 J atom^–1 ✅

👉 (ii) Radius of 5th orbit, r₅ = 13.225 Å (or 1.3225 nm) ✅

Explanation (Step-by-Step)

(i) Energy of 5th orbit
For hydrogen atom:
👉 E_n = E₁/n²

Given:
E₁ = –2.18 × 10^–18 J atom^–1
n = 5

So,
E₅ = (–2.18 × 10^–18)/5²
E₅ = (–2.18 × 10^–18)/25
E₅ = –8.72 × 10^–20 J atom^–1

📌 Negative sign shows electron is still bound to nucleus.

________________________________________

(ii) Radius of Bohr’s 5th orbit
For hydrogen:
👉 r_n = 0.529 × n² Å

For n = 5:
r₅ = 0.529 × 5² Å
r₅ = 0.529 × 25 Å
r₅ = 13.225 Å

Now convert Å to nm:
1 nm = 10 Å
So,
r₅ = 13.225/10 = 1.3225 nm

Did You Know?

💡 As n increases, orbit size increases as n², so electron gets much farther from nucleus.
💡 At the same time, energy becomes less negative, meaning electron is easier to remove from higher orbits. ⚡

Q.17: Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer

👉 The wavenumber for the longest wavelength line in Balmer series is:
1.523 × 10⁶ m⁻¹ ✅

Explanation (Step-by-Step)

📌 For Balmer series, final level is fixed at:
👉 n₁ = 2

And formula for wavenumber is:
👉 ν̄ = 1/λ = R_H[1/n₁² − 1/n₂²]
where R_H = 1.097 × 10⁷ m⁻¹.

________________________________________

For longest wavelength (λ max):
👉 wavenumber (ν̄) should be minimum.
👉 This happens for the smallest possible transition in Balmer series, i.e. n₂ = 3 to n₁ = 2.

So,
ν̄ = (1.097 × 10⁷) [1/2² − 1/3²]
ν̄ = (1.097 × 10⁷) [1/4 − 1/9]
ν̄ = (1.097 × 10⁷) [(9 − 4)/36]
ν̄ = (1.097 × 10⁷) × 5/36
ν̄ = 1.523 × 10⁶ m⁻¹

✅ Required wavenumber = 1.523 × 10⁶ m⁻¹

Did You Know?

💡 This line corresponds to the famous H_α line of hydrogen.
💡 It appears in the red region of visible spectrum (around 656 nm), and is widely used in astronomy. 🌌

Q.18: What is the energy in joules required to shift the electron of hydrogen atom from the first Bohr orbit to the fifth Bohr orbit, and what is the wavelength of light emitted when the electron returns to the ground state?
(Given: ground-state electron energy = –2.18 × 10^–11 ergs)

Answer

👉 Energy required to excite electron from n = 1 to n = 5:
ΔE = 2.09 × 10^–18 J (per atom) ✅

👉 Wavelength of light emitted when electron falls back from n = 5 to n = 1:
λ = 9.50 × 10^–8 m
= 950 Å (or 95.0 nm) ✅

Explanation (Step-by-Step)

Step 1) Convert given ground-state energy to joules
Given:
E₁ = –2.18 × 10^–11 ergs
And 1 erg = 10^–7 J, so:
E₁ = –2.18 × 10^–11 × 10^–7 J
E₁ = –2.18 × 10^–18 J

________________________________________

Step 2) Energy of 5th orbit
Using Bohr relation:
E_n = –2.18 × 10^–18 / n² J
So for n = 5:
E₅ = –2.18 × 10^–18 / 25
E₅ = –8.72 × 10^–20 J

________________________________________

Step 3) Energy required for excitation (n = 1 → n = 5)
👉 Required energy is:
ΔE = E₅ – E₁
= (–8.72 × 10^–20) – (–2.18 × 10^–18)
= 2.0928 × 10^–18 J
≈ 2.09 × 10^–18 J

________________________________________

Step 4) Wavelength emitted on return (n = 5 → n = 1)
When electron comes back, same energy is emitted as photon:
ΔE = hc/λ ⟹ λ = hc/ΔE

Take:
h = 6.626 × 10^–34 J s
c = 3.00 × 10⁸ m s^–1

So,
λ = (6.626 × 10^–34 × 3.00 × 10⁸) / (2.09 × 10^–18)
λ = 9.50 × 10^–8 m

Now convert:
• in Å: 9.50 × 10^–8 m = 950 Å
• in nm: 9.50 × 10^–8 m = 95.0 nm

Did You Know?

💡 The emitted light here is in the ultraviolet region (95 nm), not visible light.
💡 Bigger jump between levels means higher photon energy and smaller wavelength. ⚡

Q.19: The electron energy in hydrogen atom is given by E_n = (–2.18 × 10^–18)/n² J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light (in cm) that can be used to cause this transition?

Answer

👉 Energy required to remove electron from n = 2 to n = ∞:
ΔE = 5.45 × 10^–19 J ✅

👉 Longest wavelength that can cause this transition:
λ_max = 3.645 × 10^–5 cm ✅
(= 3.645 × 10^–7 m)

Explanation (Step-by-Step)

Step 1) Energy of electron in n = 2 orbit
Given:
E_n = (–2.18 × 10^–18)/n² J

So,
E₂ = (–2.18 × 10^–18)/4
E₂ = –5.45 × 10^–19 J

At n = ∞, electron is free, so:
E_∞ = 0

________________________________________

Step 2) Ionization energy from n = 2
👉 Required energy:
ΔE = E_∞ − E₂
ΔE = 0 − (−5.45 × 10^–19)
ΔE = 5.45 × 10^–19 J

________________________________________

Step 3) Longest wavelength for this transition
For threshold (minimum required energy), wavelength is maximum:
ΔE = hc/λ
⟹ λ = hc/ΔE

Using
👉 h = 6.626 × 10^–34 J s
👉 c = 3.00 × 10⁸ m s^–1

👉 λ = (6.626 × 10^–34 × 3.00 × 10⁸) / (5.45 × 10^–19)
λ = 3.645 × 10^–7 m

👉 Convert to cm:
1 m = 100 cm
λ = 3.645 × 10^–7 × 10² cm
λ = 3.645 × 10^–5 cm

✅ This is the longest wavelength that can ionize H atom from n = 2.

Did You Know?

💡 Any wavelength longer than this has lower photon energy, so it cannot ionize from n = 2.
💡 This limit is called the ionization threshold wavelength for that orbit.

Q.20: Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ m s^–1.

Answer

👉 The de Broglie wavelength of the electron is:
3.55 × 10^–11 m ✅

Explanation (Step-by-Step)

👉 Use de Broglie equation:
λ = h / mv

Where:
• h = 6.626 × 10^–34 J s
• m (electron mass) = 9.1 × 10^–31 kg
• v = 2.05 × 10⁷ m s^–1

Now substitute:
λ = (6.626 × 10^–34) / [(9.1 × 10^–31) (2.05 × 10⁷)]

First multiply denominator:
(9.1 × 2.05) × 10^–31+7
= 18.655 × 10^–24
= 1.8655 × 10^–23

Now divide:
λ = (6.626 × 10^–34) / (1.8655 × 10^–23)
= 3.55 × 10^–11 m

✅ Final wavelength = 3.55 × 10^–11 m

Did You Know?

💡 Fast-moving particles like electrons also behave like waves.
💡 This idea (matter waves) is the basis of electron microscopy, which can see very tiny structures.

Q.21: The mass of an electron is 9.1 × 10^–31 kg. If its K.E. is 3.0 × 10^–25 J, calculate its wavelength.

Answer

👉 The wavelength of the electron is:
8.97 × 10^–7 m ✅

👉 In angstrom form:
8967 Å (approximately)

Explanation (Step-by-Step)

We know:
• Mass of electron, m = 9.1 × 10^–31 kg
• Kinetic energy, K.E. = 3.0 × 10^–25 J

First, find velocity using:
K.E. = 1/2 mv²
So,
v = √(2K.E./m)

👉 v = √[(2 × 3.0 × 10^–25) / (9.1 × 10^–31)]
👉 v = √(6.593 × 10⁵)
👉 v ≈ 8.12 × 10² m s^–1

Now apply de Broglie equation:
λ = h/mv
with h = 6.626 × 10^–34 J s

👉 λ = (6.626 × 10^–34) / [(9.1 × 10^–31) (8.12 × 10²)]
👉 λ = 8.97 × 10^–7 m

So, final wavelength:
✅ λ = 8.97 × 10^–7 m

Convert to Å (1 Å = 10^–10 m):
👉 λ = 8.97 × 10^–7 / 10^–10 = 8.967 × 10³ Å
👉 λ ≈ 8967 Å

Did You Know?

💡 Lower speed means larger de Broglie wavelength.
💡 That is why, for very slow electrons, wave behavior becomes easier to observe in experiments.

Q.22: Which of the following are isoelectronic species (species having the same number of electrons)?
Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2–, Ar

Answer

👉 Na⁺ and Mg²⁺ are isoelectronic (each has 10 electrons) .

👉 K⁺, Ca²⁺, S^2–, and Ar are isoelectronic (each has 18 electrons). ✅

Explanation (Step-by-Step)

Let us count electrons in each species:

• Na (Z = 11) ⟶ Na⁺ has 11 − 1 = 10 e^–
• Mg (Z = 12) ⟶ Mg²⁺ has 12 − 2 = 10 e^–

📌 So, Na⁺ and Mg²⁺ are one isoelectronic pair.

________________________________________

• K (Z = 19) ⟶ K⁺ has 19 − 1 = 18 e^–
• Ca (Z = 20) ⟶ Ca²⁺ has 20 − 2 = 18 e^–
• S (Z = 16) ⟶ S^2– has 16 + 2 = 18 e^–
• Ar (Z = 18) ⟶ Ar has 18 e^–

📌 So, K⁺, Ca²⁺, S^2–, and Ar form another isoelectronic set.

Did You Know?

💡 Isoelectronic species have same number of electrons, but different nuclear charge (number of protons).
💡 Because of this, their sizes are different even though electron count is same.

Q.23:
(i) Write the electronic configurations of the following ions:
(a) H^– (b) Na⁺ (c) O^2– (d) F^–
(ii) What are the atomic numbers of elements whose outermost electrons are represented by:
(a) 3s¹ (b) 2p³ (c) 3p⁵
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s¹ (b) [Ne] 3s² 3p³ (c) [Ar] 4s² 3d¹

Answer

(i) Electronic configurations of ions
👉 (a) H^– : 1s²
👉 (b) Na⁺ : 1s² 2s² 2p⁶
👉 (c) O^2– : 1s² 2s² 2p⁶
👉 (d) F^– : 1s² 2s² 2p⁶ ✅

(ii) Atomic numbers
👉 (a) 3s¹ corresponds to Na, so Z = 11
👉 (b) 2p³ corresponds to N, so Z = 7
👉 (c) 3p⁵ corresponds to Cl, so Z = 17 ✅

(iii) Identify the atoms
👉 (a) [He] 2s¹ = Li (Lithium)
👉 (b) [Ne] 3s² 3p³ = P (Phosphorus)
👉 (c) [Ar] 4s² 3d¹ = Sc (Scandium) ✅

Explanation (Step-by-Step)

📌 For ions, add electrons for negative charge and remove electrons for positive charge.

• H has 1 electron, H^– has 2 → 1s²
• Na has 11 electrons, Na⁺ has 10 → neon-like
• O has 8 electrons, O^2– has 10 → neon-like
• F has 9 electrons, F^– has 10 → neon-like

📌 For outermost configurations:
• 3s¹ is alkali pattern of Na
• 2p³ is nitrogen family start (N)
• 3p⁵ is halogen pattern in period 3 (Cl)

📌 For noble gas notation:
• [He] means 2 electrons core
• [Ne] means 10 electrons core
• [Ar] means 18 electrons core
Then add remaining electrons to identify the element.

Did You Know?

💡 Na⁺, O^2–, and F^– all have 10 electrons, so they are isoelectronic with Ne.
💡 In many cases, ions become more stable after achieving a noble-gas type configuration.

Q.24: What is the lowest value of n that allows g orbitals to exist?

Answer

👉 The lowest value of principal quantum number is n = 5 ✅

Explanation (Step-by-Step)

👉 For a g orbital, azimuthal quantum number is:
l = 4

👉 Rule: for a given shell, allowed values of l are:
l = 0 to (n − 1)

So n must satisfy:
n − 1 ≥ 4
n ≥ 5

✅ Therefore, the smallest possible n for g orbital is 5 .

Did You Know?

💡 Orbital naming pattern is:
l = 0 → s, l = 1 → p, l = 2 → d, l = 3 → f, l = 4 → g
💡 In school-level chemistry, we usually use up to f orbitals, but quantum mechanically g orbitals are also possible.

Q.25: An electron is in one of the 3d orbitals. Give the possible values of n, l and m_l for this electron.

Answer

👉 For a 3d electron:
• n = 3
• l = 2
• m_l = −2, −1, 0, +1, +2 ✅

Explanation (Step-by-Step)

👉 The label 3d directly gives:
1. Principal quantum number (n) = 3
2. d-subshell means l = 2

👉 For any given l, magnetic quantum number m_l ranges from −l to +l.
So for l = 2:
m_l = −2, −1, 0, +1, +2

That gives 5 possible orientations of d orbitals.

Did You Know?

💡 A d-subshell always has 5 orbitals, and each orbital can hold 2 electrons.
💡 So total maximum electrons in any d-subshell = 10.

Q.26: An atom of an element contains 29 electrons and 35 neutrons. Deduce
(i) the number of protons and
(ii) the electronic configuration of the element.

Answer

👉 (i) Number of protons = 29 ✅

👉 (ii) Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ ✅

Explanation (Step-by-Step)

👉 In a neutral atom:
Number of protons = Number of electrons

Given electrons = 29, so protons = 29.

📌 Atomic number Z = 29, which is copper (Cu).

Now fill 29 electrons in orbitals (aufbau pattern with Cu exception stability):
• 1s²
• 2s²
• 2p⁶
• 3s²
• 3p⁶
• 3d¹⁰
• 4s¹

So final configuration:
✅ [Ar] 3d¹⁰ 4s¹

Did You Know?

💡 Copper is one of the famous exceptions in electronic configuration.
💡 It prefers 3d¹⁰ 4s¹ (fully filled d-subshell) instead of 3d⁹ 4s², because this is more stable.

Q.27: Give the number of electrons in the species: H₂⁺, H₂ and O₂⁺.

Answer

👉 H₂⁺ has 1 electron ✅

👉 H₂ has 2 electrons ✅

👉 O₂⁺ has 15 electrons ✅

Explanation (Step-by-Step)

👉 Count total electrons from atoms, then adjust for charge.

1) H₂⁺
• Each H atom has 1 electron
• H₂ would have 2 electrons
• “+” charge means 1 electron removed
So, electrons = 2 − 1 = 1

2) H₂
• 2 hydrogen atoms × 1 electron each = 2 electrons

3) O₂⁺
• Each O atom has 8 electrons
• O₂ has 16 electrons
• “+” charge means 1 electron removed
So, electrons = 16 − 1 = 15

Did You Know?

💡 H₂⁺ is the simplest molecular ion and is very important in molecular orbital theory.
💡 O₂⁺ forms when oxygen loses one electron, often in high-energy processes.

Q.28:
(i) An atomic orbital has n = 3. What are the possible values of l and m_l?
(ii) List the quantum numbers (m_l and l) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible? 1p, 2s, 2p, 3f

Answer

👉 (i) For n = 3:
Possible values of l are: 0, 1, 2
• If l = 0 ⟶ m_l = 0
• If l = 1 ⟶ m_l = −1, 0, +1
• If l = 2 ⟶ m_l = −2, −1, 0, +1, +2

________________________________________

👉 (ii) For 3d orbital:
• n = 3
• l = 2 (because d-subshell means l = 2)
• m_l = −2, −1, 0, +1, +2 ✅

________________________________________

👉 (iii) Possible orbitals from given list:
• 1p ❌ Not possible
• 2s ✅ Possible
• 2p ✅ Possible
• 3f ❌ Not possible

So, only 2s and 2p are possible.

Explanation (Step-by-Step)

📌 Rule 1: For a given n, l can be from 0 to (n−1).
So for n = 3, l = 0, 1, 2 only.

📌 Rule 2: For a given l, m_l ranges from −l to +l.

📌 Rule 3: Orbital validity check
• p orbital means l = 1, so n must be at least 2 ⟶ 1p impossible
• f orbital means l = 3, so n must be at least 4 ⟶ 3f impossible

Did You Know?

💡 The number of orbitals in a subshell is given by 2l + 1.
So for d (l = 2), number of orbitals = 5.
That is why m_l has 5 values for a d-subshell.

Q.29: Using s, p, d notations, describe the orbital with the following quantum numbers:
(a) n = 1, l = 0
(b) n = 3, l = 1
(c) n = 4, l = 2
(d) n = 4, l = 3

Answer

👉 (a) 1s orbital
👉 (b) 3p orbital
👉 (c) 4d orbital
👉 (d) 4f orbital ✅

Explanation (Step-by-Step)

Quantum number l tells the subshell type:
• l = 0 ⟶ s
• l = 1 ⟶ p
• l = 2 ⟶ d
• l = 3 ⟶ f

Now combine with given n:
• n = 1, l = 0 ⟶ 1s
• n = 3, l = 1 ⟶ 3p
• n = 4, l = 2 ⟶ 4d
• n = 4, l = 3 ⟶ 4f

Did You Know?

💡 The principal quantum number n gives shell number, while l gives subshell shape.
💡 So notation like 3p means: 3rd shell + p-subshell.

Q.30: Explain, giving reasons, which of the following sets of quantum numbers are not possible:
(a) n = 0, l = 0, m_l = 0, m_s = +1/2
(b) n = 1, l = 0, m_l = 0, m_s = −1/2
(c) n = 1, l = 1, m_l = 0, m_s = +1/2
(d) n = 2, l = 1, m_l = 0, m_s = −1/2
(e) n = 3, l = 3, m_l = −3, m_s = +1/2
(f) n = 3, l = 1, m_l = 0, m_s = +1/2

Answer (Possible / Not Possible)

👉 (a) Not possible ❌
👉 (b) Possible ✅
👉 (c) Not possible ❌
👉 (d) Possible ✅
👉 (e) Not possible ❌
👉 (f) Possible ✅

Explanation (Step-by-Step)

Use basic rules:
1) n = 1, 2, 3, ... (cannot be 0)
2) l = 0 to (n − 1)
3) m_l = −l to +l
4) m_s = +1/2 or −1/2

________________________________________

(a) n = 0, l = 0, m_l = 0, m_s = +1/2
❌ Not possible, because n cannot be 0. Minimum n is 1.

(b) n = 1, l = 0, m_l = 0, m_s = −1/2
✅ Possible.
For n = 1, only l = 0 is allowed; then m_l must be 0; m_s can be ±1/2.

(c) n = 1, l = 1, m_l = 0, m_s = +1/2
❌ Not possible.
For n = 1, l can only be 0, not 1.

(d) n = 2, l = 1, m_l = 0, m_s = −1/2
✅ Possible.
For n = 2, l can be 0 or 1. If l = 1, m_l can be −1, 0, +1; m_s valid.

(e) n = 3, l = 3, m_l = −3, m_s = +1/2
❌ Not possible.
For n = 3, l can be 0, 1, or 2 only. So l = 3 is invalid.

(f) n = 3, l = 1, m_l = 0, m_s = +1/2
✅ Possible.
All values satisfy the rules.

Did You Know?

💡 Most “invalid quantum number” questions are solved fastest by checking l range from n first.
💡 If l is wrong, the full set is automatically invalid, no need to check m_l and m_s further.

Q.31: How many electrons in an atom may have the following quantum numbers?
(a) n = 4, m_s = −1/2
(b) n = 3, l = 0

Answer

👉 (a) 16 electrons
👉 (b) 2 electrons ✅

Explanation (Step-by-Step)

(a) n = 4, m_s = −1/2
For a given shell n, maximum electrons = 2n²
So for n = 4:
Total electrons = 2 × 4² = 2 × 16 = 32

Now, spin can be +1/2 or −1/2. In a completely filled shell, half electrons have m_s = −1/2.
👉 Number with m_s = −1/2 = 32/2 = 16

________________________________________

(b) n = 3, l = 0
l = 0 means 3s subshell.
For l = 0:
• m_l = 0 (only one orbital)
• each orbital can hold 2 electrons (m_s = +1/2 and −1/2)

👉 Total electrons = 2

Did You Know?

💡 Any s-subshell (l = 0) always has only 1 orbital, so it can hold maximum 2 electrons.
💡 In general, subshell capacity = 2(2l + 1).

Q.32: Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the revolving electron.

Answer

👉 For Bohr orbit, the condition is:
2πr = nλ

So, the orbit circumference is an integer multiple (n times) of de Broglie wavelength. ✅

Explanation (Step-by-Step)

Start with Bohr’s quantization condition:
👉 mvr = nh/2π ...(1)

Rearrange:
👉 mv = nh/2πr ...(2)

Now use de Broglie relation:
👉 λ = h/mv
⇒ mv = h/λ ...(3)

Substitute (3) into (2):
h/λ = nh/2πr

Cancel h from both sides:
1/λ = n/2πr

Rearrange:
👉 2πr = nλ

Hence proved.

Did You Know?

💡 This condition explains why electrons do not spiral continuously into nucleus.
💡 Only fixed energy orbits are allowed, giving discrete line spectra of hydrogen.

Q.33: What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?

Answer

👉 The matching hydrogen transition is:
n = 2 to n = 1 (Lyman series) ✅

Explanation (Step-by-Step)

For hydrogen-like species:
👉 ν̄ = 1/λ = R_HZ²[1/n₁² − 1/n₂²]

For He⁺:
• Z = 2 (so Z² = 4)
• Given transition: n₂ = 4 to n₁ = 2

So,
ν̄(He⁺) = R_H × 4 [1/2² − 1/4²]
= R_H × 4 [1/4 − 1/16]
= R_H × 4 × (3/16)
= 3R_H/4

Now for hydrogen (Z = 1), same wavelength means same wavenumber:
R_H[1/n₁² − 1/n₂²] = 3R_H/4

So,
[1/n₁² − 1/n₂²] = 3/4

This is satisfied by:
👉 n₁ = 1, n₂ = 2
Because:
1/1² − 1/2² = 1 − 1/4 = 3/4 ✅

Hence the required hydrogen line is 2 → 1 (Lyman-α transition) .

Did You Know?

💡 He⁺ is hydrogen-like (one-electron system), so its spectrum follows the same formula with Z² factor.
💡 Because of higher nuclear charge, He⁺ lines are shifted compared with hydrogen.

Q.34: Calculate the energy required for the process
He⁺(g) → He²⁺(g) + e^–
Given: Ionization energy of H atom in ground state = 2.18 × 10^–18 J atom^–1.

Answer

👉 Energy required = 8.72 × 10^–18 J atom^–1 ✅

Explanation (Step-by-Step)

👉 He⁺ is a hydrogen-like species (only one electron).
For hydrogen-like ions in ground state (n = 1), ionization energy is proportional to Z²:

E = 2.18 × 10^–18 × Z² J atom^–1

For He⁺, nuclear charge Z = 2.
So,
E = 2.18 × 10^–18 × (2)²
E = 2.18 × 10^–18 × 4
E = 8.72 × 10^–18 J atom^–1

✅ This is the required energy for:
He⁺(g) → He²⁺(g) + e^–

Did You Know?

💡 Removing electron from He⁺ is much harder than from H, because He⁺ has higher nuclear charge and pulls electron more strongly.
💡 That is why the value is exactly 4 times hydrogen (since Z² = 4).

Q.35: If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across a scale of length 20 cm.

Answer

👉 Number of carbon atoms that can fit side by side = 1.33 × 10⁹ atoms ✅

Explanation (Step-by-Step)

👉 Given:
• Length of scale = 20 cm
• Diameter of one carbon atom = 0.15 nm

Step 1) Convert 20 cm into nm
1 cm = 10⁷ nm
So,
20 cm = 20 × 10⁷ nm
= 2 × 10⁸ nm

Step 2) Number of atoms = total length / diameter of one atom
Number of atoms = (2 × 10⁸ nm) / (0.15 nm)
= 1.333... × 10⁹

👉 Therefore,
✅ Number of carbon atoms ≈ 1.33 × 10⁹

Did You Know?

💡 Even a tiny line segment of 20 cm can hold over a billion atoms in a row.
💡 This shows how incredibly small atomic dimensions are compared to everyday objects.

Q.36: 2 × 10⁸ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Answer

👉 Radius of one carbon atom = 6.0 × 10^–9 cm
👉 In nm, radius = 0.06 nm ✅

Explanation (Step-by-Step)

Given:
• Total number of atoms, N = 2 × 10⁸
• Total length, L = 2.4 cm

If atoms are side by side, then:
👉 Diameter of one atom (d) = L / N

So,
d = (2.4 cm) / (2 × 10⁸)
d = 1.2 × 10^–8 cm

Now radius is half of diameter:
r = d/2
r = (1.2 × 10^–8)/2 cm
r = 6.0 × 10^–9 cm

________________________________________

Convert to nm:
1 cm = 10⁷ nm
r = 6.0 × 10^–9 × 10⁷ nm
r = 6.0 × 10^–2 nm
r = 0.06 nm

✅ Final radius = 6.0 × 10^–9 cm = 0.06 nm

Did You Know?

💡 In such questions, “side by side” means each atom contributes one full diameter to total length.
💡 So always use: diameter = total length / number of atoms, then radius = diameter/2.

Q.37: The diameter of zinc atom is 2.6 Å. Calculate
(a) radius of zinc atom in pm and
(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Answer

👉 (a) Radius of Zn atom = 130 pm ✅

👉 (b) Number of Zn atoms in 1.6 cm = 6.15 × 10⁷ atoms (approximately) ✅

Explanation (Step-by-Step)

(a) Radius in pm
Given diameter:
d = 2.6 Å

We know:
1 Å = 100 pm

So,
d = 2.6 × 100 = 260 pm

Radius is half of diameter:
r = d/2 = 260/2 = 130 pm
✅ Radius of zinc atom = 130 pm

________________________________________

(b) Number of atoms in 1.6 cm
If atoms are arranged side by side, each atom occupies one diameter.

Given:
• Total length, L = 1.6 cm
• Diameter of one Zn atom, d = 260 pm

Convert cm to pm:
1 cm = 10¹⁰ pm
So, 1.6 cm = 1.6 × 10¹⁰ pm

Now,
Number of atoms = Total length / Diameter of one atom
N = (1.6 × 10¹⁰ pm) / (260 pm)
N = 6.1538 × 10⁷
≈ 6.15 × 10⁷ atoms

✅ Required number of atoms = 6.15 × 10⁷

Did You Know?

💡 Unit conversion is the key in atomic-size questions.
💡 A very small atomic diameter still gives tens of millions of atoms in just a few centimeters.

Q.38: A certain particle carries 2.5 × 10^–16 C of static electric charge. Calculate the number of electrons present in it.

Answer

👉 Number of electrons present = 1560 (approximately) ✅

Explanation (Step-by-Step)

Use:
👉 Number of electrons, n = q / e

Where:
• q = 2.5 × 10^–16 C
• e = 1.602 × 10^–19 C (charge of one electron)

So,
n = (2.5 × 10^–16) / (1.602 × 10^–19)
n = (2.5 / 1.602) × 10³
n ≈ 1.56 × 10³
n ≈ 1560

✅ Therefore, the particle has about 1560 electrons worth of charge.

Did You Know?

💡 Electric charge is quantized, which means it appears in multiples of elementary charge e.
💡 That is why we divide total charge by e to find how many electrons are involved.

Q.39: In Millikan’s experiment, static electric charge on the oil drop is –1.282 × 10^–18 C. Calculate the number of electrons present on it.

Answer

👉 Number of electrons on the oil drop = 8 ✅

Explanation (Step-by-Step)

Use the relation:
👉 n = q / e

Where:
• q = –1.282 × 10^–18 C (charge on drop)
• e = –1.602 × 10^–19 C (charge of one electron)

Now calculate:
n = (–1.282 × 10^–18) / (–1.602 × 10^–19)
n = (1.282 / 1.602) × 10
n = 0.800... × 10
n = 8

✅ So, the oil drop carries charge equal to 8 extra electrons .

Did You Know?

💡 In Millikan’s oil drop experiment, charge always appeared in whole-number multiples of e.
💡 This was one of the strongest proofs that electric charge is quantized.

Q.40: In Rutherford’s experiment, heavy metal foils (like gold, platinum) are generally used for α-particle bombardment. If a thin foil of lighter atoms (like aluminium) is used, what difference would be observed?

Answer

👉 If a lighter metal foil (like Al) is used, deflection of α-particles will be much less.
👉 The number of α-particles scattered through large angles will decrease strongly.
👉 The number of α-particles that bounce back (back-scattering) will become almost negligible. ✅

Explanation (Step-by-Step)

👉 In Rutherford scattering, deflection depends on nuclear charge (positive charge in nucleus).

👉 Heavy atoms like Au have very high nuclear charge, so they repel incoming α-particles more strongly.
👉 Hence, with heavy foil, we get noticeable deflections and a few back-scattered particles.

Now if we use light atoms like Al:
📌 Nuclear charge is lower
📌 Electrostatic repulsion on α-particles is weaker
📌 So most α-particles pass through with little or no deflection

Therefore:
• small-angle scattering: reduced
• large-angle scattering: much smaller
• back scattering: nearly zero

Did You Know?

💡 Rutherford chose gold foil partly because it can be made extremely thin and has a high atomic number, both helpful for observing clear scattering effects.
💡 The rare back-scattered α-particles gave strong evidence for a tiny, dense, positively charged nucleus.

Q.41: Symbols ⁷⁹₃₅Br and ⁷⁹Br can be written, whereas ³⁵₇₉Br and ₃₅Br are not acceptable. Give a brief reason.

Answer

👉 ⁷⁹₃₅Br and ⁷⁹Br are acceptable.
👉 ³⁵₇₉Br and ₃₅Br are not acceptable. ✅

Explanation (Brief and Clear)

For nuclide notation:
• Z (atomic number) = number of protons
• A (mass number) = protons + neutrons
• Always, A ≥ Z

For bromine (Br), atomic number is fixed: Z = 35

Why these are correct:
👉 ⁷⁹₃₅Br: fully correct (A = 79, Z = 35)
👉 ⁷⁹Br: also acceptable, because Br symbol already tells Z = 35

Why these are wrong:
👉 ³⁵₇₉Br: impossible, because this means Z = 79 and A = 35, but A cannot be smaller than Z
👉 ₃₅Br: incomplete for isotope notation, because mass number (A) is missing; isotope identity is not clear

Did You Know?

💡 Element identity is decided by atomic number (Z), not mass number.
💡 Isotopes of the same element have same Z but different A, so writing A is important when specifying a particular isotope.

Q.42: An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Answer

👉 The element is bromine, and its atomic symbol is:
⁸¹₃₅Br ✅

Explanation (Step-by-Step)

Let number of protons = p
Given: neutrons are 31.7% more than protons, so
👉 Number of neutrons = 1.317p

Now mass number A is:
A = protons + neutrons
81 = p + 1.317p
81 = 2.317p

So,
p = 81 / 2.317 ≈ 34.96 ≈ 35

👉 Therefore, atomic number Z = number of protons = 35
👉 Element with Z = 35 is Br (Bromine)

Now neutrons:
n = 81 − 35 = 46

So atomic notation:
✅ ⁸¹₃₅Br

Did You Know?

💡 In isotope notation, the lower number is atomic number (protons), and upper number is mass number (protons + neutrons).
💡 Changing neutrons changes isotope, but element remains the same because protons stay the same.

Q.43: An ion with mass number 37 possesses one unit negative charge. If the ion contains 11.1% more neutrons than electrons, find the symbol of the ion.

Answer

👉 The ion is ³⁷₁₇Cl^– ✅

Explanation (Step-by-Step)

Let number of electrons in ion = e
Since ion has one negative charge:
👉 electrons = protons + 1
So, protons p = e − 1

Given neutrons are 11.1% more than electrons:
👉 neutrons n = 1.111e

Now mass number:
A = p + n = 37
So,
(e − 1) + 1.111e = 37
2.111e − 1 = 37
2.111e = 38
e ≈ 18

So:
• electrons = 18
• protons = e − 1 = 17
• neutrons = 37 − 17 = 20

Atomic number Z = 17, which is chlorine (Cl).
Hence ion symbol is:
✅ ³⁷₁₇Cl^–

Did You Know?

💡 Atomic number (protons) decides the element identity.
💡 Charge only changes electron count, not the element itself.

Q.44: An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Answer

👉 The ion is ⁵⁶₂₆Fe³⁺ ✅

Explanation (Step-by-Step)

Let number of electrons = e
Since charge is +3:
👉 protons, p = e + 3

Given neutrons are 30.4% more than electrons:
👉 neutrons, n = e + 0.304e = 1.304e

Mass number A = p + n = 56
So,
(e + 3) + 1.304e = 56
2.304e + 3 = 56
2.304e = 53
e ≈ 23

Now find protons:
p = e + 3 = 23 + 3 = 26

Atomic number Z = 26, which is Fe (iron).
Thus ion symbol is:
✅ ⁵⁶₂₆Fe³⁺

Did You Know?

💡 For cations (positive ions), electrons are fewer than protons.
💡 Fe commonly forms both Fe²⁺ and Fe³⁺, and Fe³⁺ is very common in red-brown iron compounds.

Q.45: Arrange the following radiations in increasing order of frequency:
(a) radiation from microwave oven
(b) amber light from traffic signal
(c) radiation from FM radio
(d) cosmic rays from outer space
(e) X-rays

Answer (Increasing Frequency)

👉 (c) FM radio < (a) microwave oven < (b) amber light < (e) X-rays < (d) cosmic rays ✅

Explanation (Step-by-Step)

Frequency increases as we move in the electromagnetic spectrum like this:
📌 Radio waves < Microwaves < Visible light < X-rays < Gamma/cosmic high-energy radiation

Now match items:
• FM radio → radio wave region (lowest here)
• Microwave oven → microwave region
• Amber light → visible region
• X-rays → very high frequency
• Cosmic rays → highest frequency/highest energy

So final order:
👉 FM < Microwave < Amber < X-rays < Cosmic rays

Did You Know?

💡 Frequency and wavelength are inversely related: higher frequency means shorter wavelength.
💡 Cosmic rays are extremely energetic and mostly come from high-energy events in space.

Q.46: Nitrogen laser produces radiation of wavelength 337.1 nm. If the number of photons emitted is 5.6 × 10²⁴, calculate the power of this laser.

Answer

👉 Total energy of emitted photons = 3.3 × 10⁶ J ✅

⚠️ Important: With the given data (number of photons and wavelength), we calculate total energy.
To calculate actual power, time is also required: Power = Energy / time.

Explanation (Step-by-Step)

Use photon energy relation:
👉 E_photon = hc/λ
For N photons:
👉 E_total = Nhc/λ

Given:
• N = 5.6 × 10²⁴
• h = 6.626 × 10⁻³⁴ J s
• c = 3.0 × 10⁸ m s⁻¹
• λ = 337.1 nm = 337.1 × 10⁻⁹ m

Now substitute:
E = (5.6 × 10²⁴)(6.626 × 10⁻³⁴)(3.0 × 10⁸) / (337.1 × 10⁻⁹)

First multiply numerator constants:
5.6 × 6.626 × 3.0 = 111.3168

Power of 10 in numerator:
10^24−34+8 = 10⁻²
So numerator = 111.3168 × 10⁻² = 1.113168

Now divide by 337.1 × 10⁻⁹:
E ≈ 3.30 × 10⁶ J

✅ Therefore, energy emitted = 3.3 × 10⁶ J

Did You Know?

💡 UV lasers (like 337.1 nm) have higher photon energy than visible red lasers because shorter wavelength means higher energy per photon.
💡 Even tiny wavelength changes can significantly affect photon energy.

Q.47: Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate:
(a) frequency of emission
(b) distance travelled by this radiation in 30 s
(c) energy of one quantum
(d) number of quanta present if it produces 2 J energy

Answer

👉 (a) Frequency, ν = 4.87 × 10¹⁴ s⁻¹
👉 (b) Distance in 30 s = 9.0 × 10⁹ m
👉 (c) Energy of one quantum = 3.23 × 10⁻¹⁹ J
👉 (d) Number of quanta in 2 J = 6.2 × 10¹⁸ ✅

Explanation (Step-by-Step)

Given:
• Wavelength, λ = 616 nm = 616 × 10⁻⁹ m
• Speed of light, c = 3.0 × 10⁸ m s⁻¹
• Planck constant, h = 6.626 × 10⁻³⁴ J s

________________________________________

(a) Frequency of emission
Use: ν = c/λ
ν = (3.0 × 10⁸) / (616 × 10⁻⁹)
ν = 4.87 × 10¹⁴ s⁻¹
✅ Frequency = 4.87 × 10¹⁴ s⁻¹

________________________________________

(b) Distance travelled in 30 s
Use: distance = speed × time
distance = (3.0 × 10⁸ m s⁻¹) × (30 s)
distance = 9.0 × 10⁹ m
✅ Distance = 9.0 × 10⁹ m

________________________________________

(c) Energy of one quantum
Use: E = hν = hc/λ
E = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / (616 × 10⁻⁹)
E = 3.23 × 10⁻¹⁹ J
✅ Energy per quantum = 3.23 × 10⁻¹⁹ J

________________________________________

(d) Number of quanta in 2 J
Use: N = Total energy / Energy per quantum
N = 2 / (3.23 × 10⁻¹⁹)
N ≈ 6.2 × 10¹⁸
✅ Number of quanta = 6.2 × 10¹⁸

Did You Know?

💡 616 nm lies in the orange-red region, which is why neon signs often appear bright reddish-orange.
💡 Even a small visible-light glow contains an enormous number of photons.

Q.48: In astronomical observations, signals from distant stars are generally weak. If the photon detector receives a total of 3.15 × 10^–18 J from radiation of 600 nm, calculate the number of photons received.

Answer

👉 Number of photons received, N ≈ 9.51
👉 So practically, detector receives about 10 photons (approximately) ✅

Explanation (Step-by-Step)

Use:
👉 Energy of one photon, E_photon = hc/λ

So number of photons:
👉 N = E_total / E_photon = E_total × λ / hc

Given:
• E_total = 3.15 × 10^–18 J
• λ = 600 nm = 600 × 10^–9 m
• h = 6.626 × 10^–34 J s
• c = 3.0 × 10⁸ m s^–1

Now substitute:
N = (3.15 × 10^–18 × 600 × 10^–9) / (6.626 × 10^–34 × 3.0 × 10⁸)
N ≈ 9.51

✅ Therefore, number of photons ≈ 9.5, i.e. about 10 photons.

Did You Know?

💡 In very weak astronomical signals, detectors often work in a near “photon counting” mode.
💡 This is why highly sensitive instruments are needed in astronomy, especially for distant stars and galaxies.

Q.49: Lifetimes of molecules in excited states are often measured using a pulsed radiation source of duration in nanoseconds. If the pulse duration is 2 ns and the number of photons emitted is 2.5 × 10¹⁵, calculate the energy of the source.

Answer

👉 Energy of the source = 8.28 × 10⁻¹⁰ J ✅

Explanation (Step-by-Step)

Given:
• Pulse duration, t = 2 ns = 2 × 10⁻⁹ s
• Number of photons, N = 2.5 × 10¹⁵
• Planck constant, h = 6.626 × 10⁻³⁴ J s

For this type of question, frequency is taken as:
👉 ν = 1/t = 1/(2 × 10⁻⁹) = 5 × 10⁸ s⁻¹

Energy of one photon:
👉 E_photon = hν
= (6.626 × 10⁻³⁴)(5 × 10⁸)
= 3.313 × 10⁻²⁵ J

Total energy of source:
👉 E_total = N × E_photon
= (2.5 × 10¹⁵)(3.313 × 10⁻²⁵)
= 8.2825 × 10⁻¹⁰ J
≈ 8.28 × 10⁻¹⁰ J

✅ Final energy = 8.28 × 10⁻¹⁰ J

Did You Know?

💡 Nanosecond pulses are extremely short, but they can still carry huge numbers of photons.
💡 That is why pulsed sources are very useful for studying fast molecular processes like fluorescence decay.

Q.50: The longest wavelength doublet absorption transitions are observed at 589 nm and 589.6 nm. Calculate the frequency of each transition and the energy difference between the two excited states.

Answer

👉 Frequency for 589 nm transition: 5.093 × 10¹⁴ s⁻¹
👉 Frequency for 589.6 nm transition: 5.088 × 10¹⁴ s⁻¹
👉 Energy difference between the two excited states: 3.43 × 10⁻²² J ✅

Explanation (Step-by-Step)

Given:
• λ₁ = 589 nm = 589 × 10⁻⁹ m
• λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m
• c = 3.0 × 10⁸ m s⁻¹
• h = 6.626 × 10⁻³⁴ J s

________________________________________

1) Frequency of each transition
Use: ν = c/λ

For λ₁ = 589 × 10⁻⁹ m:
ν₁ = (3.0 × 10⁸) / (589 × 10⁻⁹)
ν₁ = 5.093 × 10¹⁴ s⁻¹

For λ₂ = 589.6 × 10⁻⁹ m:
ν₂ = (3.0 × 10⁸) / (589.6 × 10⁻⁹)
ν₂ = 5.088 × 10¹⁴ s⁻¹

📌 As expected, longer wavelength (589.6 nm) gives slightly lower frequency.

________________________________________

2) Energy difference between excited states
For photon energy: E = hc/λ
So difference:
👉 ΔE = hc[(1/λ₁) − (1/λ₂)]

Substitute:
ΔE = (6.626 × 10⁻³⁴)(3.0 × 10⁸)
× [(1/(589 × 10⁻⁹)) − (1/(589.6 × 10⁻⁹))]
ΔE ≈ 3.43 × 10⁻²² J

✅ Required energy gap = 3.43 × 10⁻²² J

Did You Know?

💡 This famous close pair near 589 nm is related to sodium’s yellow doublet lines.
💡 Very small wavelength splitting means the two excited states are very close in energy.

Q.51: The work function for caesium atom is 1.9 eV. Calculate
(a) threshold wavelength and (b) threshold frequency of radiation.
If caesium is irradiated with wavelength 500 nm, calculate the kinetic energy and velocity of the ejected photoelectron.

Answer

👉 (a) Threshold wavelength, λ₀ ≈ 654 nm
👉 (b) Threshold frequency, ν₀ = 4.59 × 10¹⁴ s⁻¹
👉 For λ = 500 nm:
• Kinetic energy of photoelectron, K.E. = 9.36 × 10⁻²⁰ J
• Velocity of photoelectron, v = 4.54 × 10⁵ m s⁻¹ ✅

Explanation (Step-by-Step)

Given:
• Work function, ϕ = 1.9 eV
• 1 eV = 1.602 × 10⁻¹⁹ J

So,
ϕ = 1.9 × 1.602 × 10⁻¹⁹
ϕ = 3.0438 × 10⁻¹⁹ J

Also,
h = 6.626 × 10⁻³⁴ J s,
c = 3.0 × 10⁸ m s⁻¹,
m_e = 9.1 × 10⁻³¹ kg

________________________________________

(a) Threshold frequency
👉 ν₀ = ϕ/h
ν₀ = (3.0438 × 10⁻¹⁹) / (6.626 × 10⁻³⁴)
ν₀ = 4.59 × 10¹⁴ s⁻¹

________________________________________

(b) Threshold wavelength
👉 λ₀ = c/ν₀
λ₀ = (3.0 × 10⁸) / (4.59 × 10¹⁴)
λ₀ = 6.536 × 10⁻⁷ m
= 653.6 nm ≈ 654 nm

________________________________________

For incident wavelength λ = 500 nm
Use photoelectric equation:
👉 K.E. = hc(1/λ − 1/λ₀)

K.E. = (6.626 × 10⁻³⁴)(3.0 × 10⁸)[(1/500×10⁻⁹) − (1/654×10⁻⁹)]
K.E. ≈ 9.36 × 10⁻²⁰ J

Now velocity:
👉 K.E. = 1/2 m_ev² ⇒ v = √(2K.E./m_e)
v = √[(2 × 9.36 × 10⁻²⁰)/(9.1 × 10⁻³¹)]
v ≈ 4.54 × 10⁵ m s⁻¹

Did You Know?

💡 Caesium has a low work function, so it emits electrons even with visible light near red region.
💡 That is why alkali metals are commonly used in photoelectric devices.

Q.52: Following results are observed when sodium metal is irradiated with different wavelengths. Calculate
(a) threshold wavelength and
(b) Planck’s constant.

Given Data (Table)

λ (nm)	500	450	400
v × 10⁵ (cm s⁻¹)	2.55	4.35	5.35

Answer

👉 (a) Threshold wavelength, λ₀ ≈ 531 nm ✅

👉 (b) Planck’s constant, h ≈ 6.6 × 10⁻³⁴ J s (experimental value, close estimate) ✅

Explanation (Step-by-Step)

For photoelectric effect:
👉 hν = hν₀ + (1/2)mv²
👉 hc(1/λ − 1/λ₀) = (1/2)mv²

Using the three observations and eliminating constants by ratio method (as in standard NCERT approach), we get:
✅ λ₀ ≈ 531 nm

Now use one observation (for example λ = 400 nm, v = 5.35 × 10⁵ m s⁻¹) in:
👉 h = [(1/2)mv²] / [c(1/λ − 1/λ₀)]

Substituting m = 9.1 × 10⁻³¹ kg, c = 3.0 × 10⁸ m s⁻¹, λ₀ = 531 nm gives an experimental value close to:
✅ h ≈ 6.6 × 10⁻³⁴ J s

Did You Know?

💡 In real experiments, calculated h may differ slightly from 6.626 × 10⁻³⁴ J s because of measurement errors and surface effects of metal.
💡 The threshold wavelength is directly linked to work function: larger λ₀ means lower work function.

Q.53: In a photoelectric experiment with silver, photoelectrons are stopped by applying 0.35 V when radiation of 256.7 nm is used. Calculate the work function of silver.

Answer

👉 Work function of silver, ϕ = 4.48 eV ✅
(approximately 7.18 × 10⁻¹⁹ J )

Explanation (Step-by-Step)

Use photoelectric equation:
👉 hν = w₀ + K.E._max

Given stopping potential V₀ = 0.35 V, so
👉 K.E._max = eV₀ = 0.35 eV

Now find photon energy from wavelength λ = 256.7 nm:
👉 E = hc/λ
In eV form, convenient relation:
E (eV) = 1240 / λ(nm)

So,
E = 1240 / 256.7 = 4.83 eV (approx)

Now:
w₀ = E − K.E._max
w₀ = 4.83 − 0.35
w₀ = 4.48 eV

✅ Work function of Ag = 4.48 eV

________________________________________

Quick Joule conversion
1 eV = 1.602 × 10⁻¹⁹ J
w₀ = 4.48 × 1.602 × 10⁻¹⁹
= 7.18 × 10⁻¹⁹ J (approx)

Did You Know?

💡 Stopping potential directly tells maximum kinetic energy in eV (numerically same value).
💡 Silver has relatively high work function, so it needs high-frequency (UV) light for strong photoemission.

Q.54: If a photon of wavelength 150 pm strikes an atom and one inner-bound electron is ejected with velocity 1.5 × 10⁷ m s⁻¹, calculate the energy with which it was bound to the nucleus.

Answer

👉 Binding energy of the electron = 1.22 × 10⁻¹⁵ J
👉 In eV, this is 7.63 × 10³ eV (≈ 7.63 keV) ✅

Explanation (Step-by-Step)

Use energy conservation in photoejection:
👉 Photon energy = Binding energy + Kinetic energy
So,
👉 E_binding = E_photon − K.E.

________________________________________

1) Photon energy from wavelength
Given:
λ = 150 pm = 150 × 10⁻¹² m = 1.5 × 10⁻¹⁰ m
Formula:
👉 E_photon = hc/λ
E_photon = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / (1.5 × 10⁻¹⁰)
= 1.3252 × 10⁻¹⁵ J

________________________________________

2) Kinetic energy of ejected electron
Given:
m = 9.1 × 10⁻³¹ kg, v = 1.5 × 10⁷ m s⁻¹
👉 K.E. = (1/2)mv²
= (1/2)(9.1 × 10⁻³¹)(1.5 × 10⁷)²
= 1.02375 × 10⁻¹⁶ J

________________________________________

3) Binding energy
👉 E_binding = E_photon − K.E.
= (1.3252 × 10⁻¹⁵) − (1.02375 × 10⁻¹⁶)
= 1.222825 × 10⁻¹⁵ J
≈ 1.22 × 10⁻¹⁵ J

Convert to eV:
👉 E(eV) = (1.222825 × 10⁻¹⁵) / (1.602 × 10⁻¹⁹)
≈ 7.63 × 10³ eV

✅ Binding energy = 1.22 × 10⁻¹⁵ J = 7.63 keV

Did You Know?

💡 A binding energy in keV range indicates this is an inner-shell electron (like K/L shell region), not a loosely bound outer electron.
💡 X-ray photons (very short wavelength like 150 pm) are energetic enough to knock out such inner electrons.

Q.55: Emission transitions in Paschen series end at n = 3 and are represented by
ν = 3.29 × 10¹⁵ [1/3² − 1/n²] Hz.
If transition is observed at 1285 nm, calculate n and identify the spectral region.

Answer

👉 The upper orbit is n = 5 ✅
👉 This line lies in the infrared (IR) region ✅

Explanation (Step-by-Step)

Given wavelength:
λ = 1285 nm = 1.285 × 10⁻⁶ m

First find frequency:
👉 ν = c/λ
= (3.0 × 10⁸) / (1.285 × 10⁻⁶)
= 2.3346 × 10¹⁴ s⁻¹

Now use Paschen formula:
ν = 3.29 × 10¹⁵ [1/9 − 1/n²]

2.3346 × 10¹⁴ = 3.29 × 10¹⁵ [1/9 − 1/n²]

Divide both sides by 3.29 × 10¹⁵:
(2.3346 × 10¹⁴) / (3.29 × 10¹⁵) = 1/9 − 1/n²
0.071 ≈ 0.111 − 1/n²

So,
1/n² = 0.111 − 0.071 = 0.040
n² = 1/0.040 = 25
n = 5

✅ Transition is 5 → 3 (Paschen series)
Since Paschen lines are in IR, this wavelength is in:
✅ Infrared region

Did You Know?

💡 Hydrogen spectral series: Lyman (UV), Balmer (visible), Paschen (IR).
💡 So whenever final level is n = 3, you can directly expect infrared emission.

Q.56: Calculate the wavelength for an emission transition that starts from orbit radius 1.3225 nm and ends at 211.6 pm. Name the series and spectral region.

Answer

👉 Wavelength of emitted radiation = 434 nm (approximately) ✅
👉 This transition belongs to the Balmer series ✅
👉 It lies in the visible region ✅

Explanation (Step-by-Step)

For hydrogen-like atom:
👉 r_n = (52.9 n²/Z) pm
For same atom/species, r ∝ n².

Given:
• r_start = 1.3225 nm = 1322.5 pm
• r_end = 211.6 pm

So,
r_start/r_end = n_start²/n_end²
1322.5 / 211.6 = 6.25

So, n_start/n_end = √6.25 = 2.5
Small integer pair matching 2.5 is:
👉 n_start = 5, n_end = 2
So transition is 5 → 2.

________________________________________

Now use Rydberg formula:
👉 ν̄ = R_H(1/n₁² − 1/n₂²)
with n₁ = 2, n₂ = 5, R_H = 1.097 × 10⁷ m⁻¹

ν̄ = 1.097 × 10⁷(1/4 − 1/25)
= 1.097 × 10⁷(21/100)
= 2.3037 × 10⁶ m⁻¹

Then,
λ = 1/ν̄ = 1/(2.3037 × 10⁶) m
= 4.34 × 10⁻⁷ m
= 434 nm

✅ Hence wavelength = 434 nm , Balmer line, visible region.

Did You Know?

💡 Balmer transitions are those where electron ends at n = 2, and many of these lines are visible to the human eye.
💡 The 5 → 2 transition is the well-known H-γ line.

Q.57: If the velocity of an electron in an electron microscope is 1.6 × 10⁶ m s⁻¹, calculate its de Broglie wavelength.

Answer

👉 de Broglie wavelength of the electron = 4.55 × 10⁻¹⁰ m
= 0.455 nm
= 455 pm ✅

Explanation (Step-by-Step)

Use de Broglie equation:
👉 λ = h / mv

Given:
• h = 6.626 × 10⁻³⁴ J s
• m_e = 9.1 × 10⁻³¹ kg
• v = 1.6 × 10⁶ m s⁻¹

Substitute:
λ = (6.626 × 10⁻³⁴) / [(9.1 × 10⁻³¹)(1.6 × 10⁶)]

First denominator:
9.1 × 1.6 = 14.56
⇒ denominator = 14.56 × 10⁻²⁵ = 1.456 × 10⁻²⁴

Now divide:
λ = (6.626 × 10⁻³⁴) / (1.456 × 10⁻²⁴)
= 4.55 × 10⁻¹⁰ m

✅ Final wavelength:
• 4.55 × 10⁻¹⁰ m
• 0.455 nm
• 455 pm

Did You Know?

💡 Electron microscopes can resolve much smaller details than light microscopes because electron wavelengths are much shorter than visible light wavelengths.
💡 That is why they are widely used to study viruses, proteins, and nanomaterials.

Q.58: In a neutron diffraction microscope, if wavelength used is 800 pm, calculate the characteristic velocity of neutron.

Answer

👉 Characteristic velocity of neutron = 494 m s⁻¹ ✅

Explanation (Step-by-Step)

Use de Broglie relation:
👉 λ = h/mv ⟹ v = h/(mλ)

Given:
• λ = 800 pm = 800 × 10⁻¹² m = 8 × 10⁻¹⁰ m
• h = 6.626 × 10⁻³⁴ J s
• m (neutron) = 1.675 × 10⁻²⁷ kg

Substitute:
v = (6.626 × 10⁻³⁴) / [(1.675 × 10⁻²⁷)(8 × 10⁻¹⁰)]

Denominator:
1.675 × 8 = 13.4
and 10⁻²⁷ × 10⁻¹⁰ = 10⁻³⁷
So denominator = 13.4 × 10⁻³⁷

Now:
v = (6.626/13.4) × 10³
v = 0.494 × 10³
v = 494 m s⁻¹

✅ Final velocity = 494 m s⁻¹

Did You Know?

💡 Neutron diffraction is very useful for locating light atoms (especially hydrogen) in crystal structures.
💡 Unlike electrons, neutrons are uncharged, so they can penetrate deeply into materials.

Q.59: If the velocity of the electron in Bohr’s first orbit is 2.19 × 10⁶ m s⁻¹, calculate the de Broglie wavelength associated with it.

Answer

👉 de Broglie wavelength, λ = 3.324 × 10⁻¹⁰ m
= 0.3324 nm
= 332.4 pm ✅

Explanation (Step-by-Step)

Use de Broglie equation:
👉 λ = h / mv

Given:
• h = 6.626 × 10⁻³⁴ J s
• m = 9.1 × 10⁻³¹ kg
• v = 2.19 × 10⁶ m s⁻¹

Substitute:
λ = (6.626 × 10⁻³⁴) / [(9.1 × 10⁻³¹)(2.19 × 10⁶)]

First denominator:
9.1 × 2.19 = 19.929
and 10⁻³¹ × 10⁶ = 10⁻²⁵
So denominator = 19.929 × 10⁻²⁵

Now divide:
λ = (6.626 / 19.929) × 10⁻⁹ m
λ = 0.3324 × 10⁻⁹ m
λ = 3.324 × 10⁻¹⁰ m

✅ Therefore, de Broglie wavelength = 332.4 pm (approx).

Did You Know?

💡 In Bohr’s first orbit, this wavelength fits the standing-wave condition 2πr = nλ (with n = 1).
💡 This is a neat link between Bohr’s model and de Broglie’s wave model.

Q.60: The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 10⁵ m s⁻¹. If a hockey ball of mass 0.1 kg moves with this velocity, calculate its associated wavelength.

Answer

👉 de Broglie wavelength of hockey ball = 1.52 × 10⁻³⁸ m ✅

Explanation (Step-by-Step)

Use de Broglie equation:
👉 λ = h / mv

Given:
• h = 6.626 × 10⁻³⁴ J s
• m = 0.1 kg
• v = 4.37 × 10⁵ m s⁻¹

Substitute:
λ = (6.626 × 10⁻³⁴) / [(0.1)(4.37 × 10⁵)]

First denominator:
(0.1)(4.37 × 10⁵) = 4.37 × 10⁴

Now:
λ = (6.626 × 10⁻³⁴) / (4.37 × 10⁴)
= (6.626/4.37) × 10⁻³⁸
= 1.516 × 10⁻³⁸ m

Rounded:
✅ λ ≈ 1.52 × 10⁻³⁸ m

Did You Know?

💡 This wavelength is unimaginably tiny, so wave nature of everyday objects (like balls) cannot be observed experimentally.
💡 de Broglie effects become noticeable mainly for microscopic particles (electrons, neutrons, atoms).

Q.61: If the position of an electron is measured within an accuracy of ±0.002 nm, calculate the uncertainty in momentum. If momentum is taken as h/(4π × 0.05 nm), is there any problem in defining this value?

Answer

👉 Uncertainty in momentum, Δp = 2.64 × 10⁻²³ kg m s⁻¹
👉 Given momentum value, p = 1.055 × 10⁻²⁴ kg m s⁻¹
👉 Since p << Δp, this momentum cannot be defined meaningfully with that position accuracy. ✅

Explanation (Step-by-Step)

Use Heisenberg uncertainty principle:
👉 Δx·Δp ≥ h/4π

Given:
Δx = 0.002 nm = 0.002 × 10⁻⁹ m = 2.0 × 10⁻¹² m

So minimum uncertainty in momentum:
👉 Δp = h/(4πΔx)
Δp = (6.626 × 10⁻³⁴) / [4 × 3.142 × (2.0 × 10⁻¹²)]
Δp = 2.638 × 10⁻²³ kg m s⁻¹
≈ 2.64 × 10⁻²³ kg m s⁻¹

________________________________________

Now given momentum:
👉 p = h/(4π × 0.05 nm)
0.05 nm = 5 × 10⁻¹¹ m
p = (6.626 × 10⁻³⁴) / [4 × 3.142 × 5 × 10⁻¹¹]
p = 1.055 × 10⁻²⁴ kg m s⁻¹

Compare:
• Δp = 2.64 × 10⁻²³
• p = 1.055 × 10⁻²⁴

Clearly, Δp is about 25 times larger than p.
✅ So yes, there is a problem: with this measurement accuracy, the momentum value is too uncertain to be physically well-defined.

Did You Know?

💡 If uncertainty in a quantity is larger than the quantity itself, the value loses practical meaning.
💡 This is why electron behavior must be described probabilistically, not like a classical particle with exact path and momentum.

Q.62: The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. Also identify combinations having same energy.
1. n = 4, l = 2, m_l = −2, m_s = −1/2
2. n = 3, l = 2, m_l = +1, m_s = +1/2
3. n = 4, l = 1, m_l = 0, m_s = +1/2
4. n = 3, l = 2, m_l = −2, m_s = −1/2
5. n = 3, l = 1, m_l = −1, m_s = +1/2
6. n = 4, l = 1, m_l = 0, m_s = +1/2

Answer

👉 (increasing energy): (5) < (2) = (4) < (3) = (6) < (1) ✅

Explanation (Step-by-Step)

First map each set to subshell:
• (1) n=4, l=2 → 4d
• (2) n=3, l=2 → 3d
• (3) n=4, l=1 → 4p
• (4) n=3, l=2 → 3d
• (5) n=3, l=1 → 3p
• (6) n=4, l=1 → 4p

Now apply (n + l) rule (for multi-electron atoms):
• 3p: n+l = 3+1 = 4
• 3d: n+l = 3+2 = 5
• 4p: n+l = 4+1 = 5
• 4d: n+l = 4+2 = 6

Lower (n+l) → lower energy
If (n+l) is same, lower n has lower energy.

So:
• 3p lowest
• between 3d and 4p (both 5), 3d < 4p
• then 4d highest among these

Hence:
👉 3p < 3d = 3d < 4p = 4p < 4d
i.e.
👉 (5) < (2) = (4) < (3) = (6) < (1)

Did You Know?

💡 In absence of magnetic field, energy does not depend on m_l and m_s for same subshell, so 3d electrons here are degenerate, and 4p electrons are degenerate.
💡 That is why pairs (2,4) and (3,6) have equal energies.

Q.63: Bromine atom has 35 electrons. It contains 6 electrons in 2p, 6 electrons in 3p and 5 electrons in 4p. Which of these electrons experiences the lowest effective nuclear charge?

Answer

👉 The 4p electron experiences the lowest effective nuclear charge (Z_eff) ✅

Explanation (Step-by-Step)

Effective nuclear charge depends on:
👉 Actual nuclear charge (attraction by nucleus)
👉 Shielding/screening by inner electrons

Now compare 2p, 3p and 4p electrons:
• 2p is close to nucleus, least shielded → higher Z_eff
• 3p is farther, more shielded than 2p
• 4p is farthest from nucleus and highly shielded by inner shells (1s, 2s/2p, 3s/3p/3d)

So attraction felt by 4p electron is minimum.
✅ Therefore, 4p electron has lowest effective nuclear charge .

Did You Know?

💡 Lower Z_eff usually means electron is easier to remove.
💡 That is why outermost electrons mainly decide chemical reactivity.

Q.64: Among the following pairs of orbitals, which orbital experiences larger effective nuclear charge?
(i) 2s and 3s
(ii) 4d and 4f
(iii) 3d and 3p

Answer

👉 (i) 2s
👉 (ii) 4d
👉 (iii) 3p ✅

Explanation (Step-by-Step)

Rule to use:
👉 Orbital with greater penetration (closer average approach to nucleus) feels higher effective nuclear charge (Z_eff)

General penetration trend:
s > p > d > f
Also, within same type, lower n is usually more penetrating.

Now apply:

(i) 2s vs 3s
Both are s orbitals, but 2s is closer to nucleus than 3s.
✅ So 2s feels higher Z_eff.

(ii) 4d vs 4f
For same n = 4, penetration: d > f.
✅ So 4d feels higher Z_eff.

(iii) 3d vs 3p
For same n = 3, penetration: p > d.
✅ So 3p feels higher Z_eff.

Did You Know?

💡 Because s orbitals penetrate most, s-electrons are usually held more tightly than p, d, f electrons of the same shell.
💡 This penetration idea helps explain many periodic trends and electron filling behavior.

Q.65: The unpaired electrons in Al and Si are present in 3p orbital. Which electron will experience more effective nuclear charge from the nucleus?

Answer

👉 The unpaired 3p electron in Si experiences more effective nuclear charge than in Al. ✅

Explanation (Step-by-Step)

Electronic configurations:
• Al (Z = 13): [Ne] 3s² 3p¹
• Si (Z = 14): [Ne] 3s² 3p²

Both unpaired electrons are in the same type of orbital (3p), so shielding pattern is quite similar.

But Si has:
👉 one extra proton in nucleus (higher Z)
👉 almost similar shielding compared to Al

So net attraction (effective nuclear charge, Z_eff) is higher for Si 3p electron.
✅ Therefore, Si’s unpaired 3p electron feels more effective nuclear charge .

Did You Know?

💡 Across a period (left to right), Z_eff generally increases.
💡 That is why atomic size decreases from Al to Si and ionization tendency changes gradually.

Q.66: Indicate the number of unpaired electrons in:
(a) P
(b) Si
(c) Cr
(d) Fe and
(e) Kr

Answer

👉 (a) P: 3 unpaired electrons
👉 (b) Si: 2 unpaired electrons
👉 (c) Cr: 6 unpaired electrons
👉 (d) Fe: 4 unpaired electrons
👉 (e) Kr: 0 unpaired electrons ✅

Explanation (Step-by-Step)

(a) P (Z = 15)
Configuration: [Ne] 3s² 3p³
In 3p, three electrons occupy three different p orbitals singly (Hund’s rule).
✅ Unpaired = 3

(b) Si (Z = 14)
Configuration: [Ne] 3s² 3p²
Two electrons occupy two separate p orbitals.
✅ Unpaired = 2

(c) Cr (Z = 24)
Configuration: [Ar] 3d⁵ 4s¹ (exception)
3d⁵ gives 5 unpaired + 4s¹ gives 1 unpaired.
✅ Unpaired = 6

(d) Fe (Z = 26)
Configuration: [Ar] 3d⁶ 4s²
In 3d⁶, five d orbitals fill singly first, then one pairs.
So 4 electrons remain unpaired in 3d.
✅ Unpaired = 4

(e) Kr (Z = 36)
Configuration: [Ar] 3d¹⁰ 4s² 4p⁶
All subshells filled completely.
✅ Unpaired = 0

Did You Know?

💡 Number of unpaired electrons helps predict magnetic behavior.
💡 More unpaired electrons generally means stronger paramagnetism.

Q.67:
(a) How many subshells are associated with n = 4?
(b) How many electrons will be present in these subshells having m_s = −1/2 for n = 4?

Answer

👉 (a) Number of subshells for n = 4 is 4 ✅
👉 (b) Number of electrons with m_s = −1/2 is 16 ✅

Explanation (Step-by-Step)

(a) Number of subshells for n = 4
For a given n, l values are 0 to (n−1).
So for n = 4:
• l = 0 (4s)
• l = 1 (4p)
• l = 2 (4d)
• l = 3 (4f)
Total subshells = 4

________________________________________

(b) Electrons with m_s = −1/2 in n = 4 shell
Total number of orbitals in shell n is:
👉 n² = 4² = 16 orbitals
Each orbital can contain one electron with spin −1/2 (and one with +1/2).
So maximum electrons with m_s = −1/2 in n = 4 shell:
👉 16
✅ Final: 16 electrons

Did You Know?

💡 Maximum total electrons in n = 4 shell is 2n² = 32.
💡 Exactly half of them can have m_s = +1/2 and half m_s = −1/2 when the shell is fully filled.

NCERT Solutions for Class 11 Chemistry Chapter 2 – Atomic Structure

Class 11 Chemistry Chapter 2 – NCERT Solutions (Structure of the Atom)

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