Question:medium

An electron in the ground state of a hydrogen atom absorbs 12.09 eV energy. The angular momentum of the electron increases by

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It is helpful to memorize the energies of the first few levels of the hydrogen atom: E1 = -13.6 eV, E2 = -3.4 eV, E3 = -1.51 eV, E4 = -0.85 eV. This can save time in calculating the final state.
Updated On: Mar 27, 2026
  • (h/2\(\pi\))
  • 2(h/2\(\pi\))
  • 3(h/2\(\pi\))
  • 4(h/2\(\pi\))
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The Correct Option is B

Solution and Explanation


Step 1: Conceptual Framework:
This problem integrates hydrogen atom energy level transitions with angular momentum quantization based on the Bohr model. The objective is to determine the electron's final energy level post-absorption and subsequently calculate the alteration in its angular momentum.

Step 2: Relevant Formulas/Methodology:
1. Hydrogen atom electron energy in the \(n\)-th orbit: \(E_n = -\frac{13.6}{n^2}\) eV. 2. Electron angular momentum in the \(n\)-th orbit: \(L_n = n \frac{h}{2\pi} = n\hbar\). 3. Energy change: \(\Delta E = E_{final} - E_{initial}\). 4. Angular momentum change: \(\Delta L = L_{final} - L_{initial}\).

Step 3: Detailed Analysis:
Part 1: Determine the final energy level (\(n_f\)).
The electron begins in the ground state, meaning its initial principal quantum number is \(n_i = 1\). The initial energy is \(E_1 = -\frac{13.6}{1^2} = -13.6\) eV. The electron absorbs 12.09 eV. The final energy is \(E_f = E_i + \Delta E = -13.6 \, \text{eV} + 12.09 \, \text{eV} = -1.51 \, \text{eV}\). Using the energy formula, the final quantum number \(n_f\) is calculated: \[ E_f = -\frac{13.6}{n_f^2} \] \[ -1.51 = -\frac{13.6}{n_f^2} \] \[ n_f^2 = \frac{13.6}{1.51} \approx 9 \] \[ n_f = 3 \] Consequently, the electron transitions from the n=1 state to the n=3 state.
Part 2: Compute the change in angular momentum (\(\Delta L\)).
Initial angular momentum (\(n_i = 1\)): \[ L_i = n_i \frac{h}{2\pi} = 1 \cdot \frac{h}{2\pi} \] Final angular momentum (\(n_f = 3\)): \[ L_f = n_f \frac{h}{2\pi} = 3 \cdot \frac{h}{2\pi} \] The increment in angular momentum is: \[ \Delta L = L_f - L_i = 3\frac{h}{2\pi} - 1\frac{h}{2\pi} = 2\frac{h}{2\pi} \]

Step 4: Conclusion:
The electron's angular momentum increases by 2(h/2\(\pi\)).

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