JEE MainPhysics

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Change in Electron Angular Momentum During Hydrogen Excitation

Q: Why is the change in angular momentum just h/(2π)?

Because the electron transitions from n=1 to n=2, so Δ n = 2 - 1 = 1. The change Δ L = Δ n · h/2π = 1 · h/2π.

A hydrogen atom in is ground state absorbs 10.2eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of: (Given, Plank's constant = 6.6× 10^(-34)Js)

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A2.10× 10^(-34)JsB1.05× 10^(-34)JsC3.15× 10^(-34)JsD4.2× 10^(-34)Js

Step-by-step solutionView

Correct answer

Excitation from $n = 1$ to $n = 2$ increases the principal quantum number by $Δ n = 1$ , resulting in an angular momentum increase of $\frac{h}{2 π} \approx 1.05 \times 10^{- 34} Js$ .

Option analysis

Why each option works or fails

A · 2.10× 10^(-34)Js

The student calculates the total final angular momentum $L_{2} = 2 ℏ \approx 2.10 \times 10^{- 34} Js$ instead of the increase $Δ L = L_{2} - L_{1}$ . Identify whether the problem asks for the final value or the change (increase) in angular momentum: $Δ L = (n_{2} - n_{1}) \frac{h}{2 π}$ .

B · 1.05× 10^(-34)Js

Correct option: Absorbing $10.2 eV$ promotes the electron from $n = 1$ ( $- 13.6 eV$ ) to $n = 2$ ( $- 3.4 eV$ ), so $Δ n = 1$ and $Δ L = 1 \times \frac{h}{2 π} = \frac{6.6 \times 10^{- 34}}{2 \times 3.14} \approx 1.05 \times 10^{- 34} Js$ . None needed.

C · 3.15× 10^(-34)Js

The student misidentifies the final level as $n = 3$ or assumes the transition is from $n = 1$ to $n = 4$ with $Δ n = 3$ , giving $Δ L = 3 ℏ \approx 3.15 \times 10^{- 34} Js$ . Use $E_{n} = - \frac{13.6}{n^{2}} eV$ to verify that $E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2 eV$ , which confirms $n_{2} = 2$ , not $n = 3$ or $n = 4$ .

D · 4.2× 10^(-34)Js

The student mistakes the change in energy or quantum numbers as scaling with $n^{2}$ , leading to a multiplier of $4 ℏ \approx 4.2 \times 10^{- 34} Js$ . Angular momentum in the Bohr model scales linearly with $n$ via $L = \frac{n h}{2 π}$ , not quadratically.

Reviewed route

Solution

StepWorking

01Given

Initial state: ground state ( $n_{1} = 1$ ), absorbed energy $Δ E = 10.2 eV$ , Planck's constant $h = 6.6 \times 10^{- 34} J s$ .

02Find

Change in orbital angular momentum of the electron: $Δ L = L_{f} - L_{i}$ .

03Visualise

The electron absorbs energy $10.2 eV$ in the ground state ( $E_{1} = - 13.6 eV$ ), transitioning to an excited state with energy $E_{n} = - 13.6 + 10.2 = - 3.4 eV$ , corresponding to $n = 2$ .

04Strategise

Find principal quantum number $n_{2}$ using $Δ E = 13.6 (1 - 1 / n^{2}) eV$ . Then apply Bohr's postulate for angular momentum $L = \frac{n h}{2 π}$ , giving $Δ L = (n_{2} - n_{1}) \frac{h}{2 π}$ .

05Execute

$13.6 (1 - 1 / n^{2}) = 10.2 ⟹ 1 - 1 / n^{2} = 0.75 ⟹ 1 / n^{2} = 0.25 ⟹ n = 2$ . Thus, $Δ n = 2 - 1 = 1$ . The change in angular momentum is $Δ L = \frac{h}{2 π} = \frac{6.6 \times 10^{- 34}}{2 \times (22 / 7)} = 1.05 \times 10^{- 34} J s$ .

✓Verify

Units of angular momentum are $J s$ , consistent with $h$ . The factor $ℏ = h / (2 π) \approx 1.055 \times 10^{- 34} J s$ , matching $1.05 \times 10^{- 34} J s$ .

Hints that build this answer step by step

To which principal quantum number level $n_{2}$ is the electron excited when it absorbs $10.2 eV$ from the ground state ( $n_{1} = 1$ )?
$n_{2} = 2$
What is the increase in orbital angular momentum $Δ L$ for this transition?
$Δ L = (2 - 1) \frac{h}{2 π} = \frac{6.6 \times 10^{- 34}}{2 π} \approx 1.05 \times 10^{- 34} Js$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why is the change in angular momentum just

h / (2 π)

Because the electron transitions from $n = 1$ to $n = 2$ , so $Δ n = 2 - 1 = 1$ . The change $Δ L = Δ n \cdot \frac{h}{2 π} = 1 \cdot \frac{h}{2 π}$ .

Answer

Excitation from $n = 1$ to $n = 2$ increases the principal quantum number by $Δ n = 1$ , resulting in an angular momentum increase of $\frac{h}{2 π} \approx 1.05 \times 10^{- 34} Js$ .

Why each option works or fails

A: 2.10× 10^(-34)Js - The student calculates the total final angular momentum $L_{2} = 2 ℏ \approx 2.10 \times 10^{- 34} Js$ instead of the increase $Δ L = L_{2} - L_{1}$ . Identify whether the problem asks for the final value or the change (increase) in angular momentum: $Δ L = (n_{2} - n_{1}) \frac{h}{2 π}$ .
B · correct: 1.05× 10^(-34)Js - Correct option: Absorbing $10.2 eV$ promotes the electron from $n = 1$ ( $- 13.6 eV$ ) to $n = 2$ ( $- 3.4 eV$ ), so $Δ n = 1$ and $Δ L = 1 \times \frac{h}{2 π} = \frac{6.6 \times 10^{- 34}}{2 \times 3.14} \approx 1.05 \times 10^{- 34} Js$ . None needed.
C: 3.15× 10^(-34)Js - The student misidentifies the final level as $n = 3$ or assumes the transition is from $n = 1$ to $n = 4$ with $Δ n = 3$ , giving $Δ L = 3 ℏ \approx 3.15 \times 10^{- 34} Js$ . Use $E_{n} = - \frac{13.6}{n^{2}} eV$ to verify that $E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2 eV$ , which confirms $n_{2} = 2$ , not $n = 3$ or $n = 3$ .
D: 4.2× 10^(-34)Js - The student mistakes the change in energy or quantum numbers as scaling with $n^{2}$ , leading to a multiplier of $4 ℏ \approx 4.2 \times 10^{- 34} Js$ . Angular momentum in the Bohr model scales linearly with $n$ via $L = \frac{n h}{2 π}$ , not quadratically.

Step-by-step solution

Given: Initial state: ground state ( $n_{1} = 1$ ), absorbed energy $Δ E = 10.2 eV$ , Planck's constant $h = 6.6 \times 10^{- 34} J s$ .
Find: Change in orbital angular momentum of the electron: $Δ L = L_{f} - L_{i}$ .
Visualise: The electron absorbs energy $10.2 eV$ in the ground state ( $E_{1} = - 13.6 eV$ ), transitioning to an excited state with energy $E_{n} = - 13.6 + 10.2 = - 3.4 eV$ , corresponding to $n = 2$ .
Strategise: Find principal quantum number $n_{2}$ using $Δ E = 13.6 (1 - 1 / n^{2}) eV$ . Then apply Bohr's postulate for angular momentum $L = \frac{n h}{2 π}$ , giving $Δ L = (n_{2} - n_{1}) \frac{h}{2 π}$ .
Execute: $13.6 (1 - 1 / n^{2}) = 10.2 ⟹ 1 - 1 / n^{2} = 0.75 ⟹ 1 / n^{2} = 0.25 ⟹ n = 2$ . Thus, $Δ n = 2 - 1 = 1$ . The change in angular momentum is $Δ L = \frac{h}{2 π} = \frac{6.6 \times 10^{- 34}}{2 \times (22 / 7)} = 1.05 \times 10^{- 34} J s$ .
Verify: Units of angular momentum are $J s$ , consistent with $h$ . The factor $ℏ = h / (2 π) \approx 1.055 \times 10^{- 34} J s$ , matching $1.05 \times 10^{- 34} J s$ .

Change in Electron Angular Momentum During Hydrogen Excitation

What feels right?

Excitation from $n = 1$ to $n = 2$ increases the principal quantum number by $Δ n = 1$ , resulting in an angular momentum increase of $\frac{h}{2 π} \approx 1.05 \times 10^{- 34} Js$ .

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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