Question

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

• A. (h/2\(\pi\))
• B. 2(h/2\(\pi\))
• C. 3(h/2\(\pi\))
• D. 4(h/2\(\pi\))

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem connects the energy level transitions in a hydrogen atom with the quantization of angular momentum as described by the Bohr model. We first need to identify the final energy level of the electron after absorbing the given energy, and then calculate the change in its angular momentum.

Step 2: Key Formula or Approach:
1. Energy of an electron in the \(n\)-th orbit of a hydrogen atom: \(E_n = -\frac{13.6}{n^2}\) eV. 2. Angular momentum of an electron in the \(n\)-th orbit: \(L_n = n \frac{h}{2\pi} = n\hbar\). 3. The change in energy is \(\Delta E = E_{final} - E_{initial}\). 4. The change in angular momentum is \(\Delta L = L_{final} - L_{initial}\).

Step 3: Detailed Explanation:
Part 1: Find the final energy level (n\(_f\)).
The electron is initially in the ground state, so 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 of energy. The final energy of the electron is \(E_f = E_i + \Delta E = -13.6 \, \text{eV} + 12.09 \, \text{eV} = -1.51 \, \text{eV}\). Now, we find the final quantum number \(n_f\) using the energy formula: \[ E_f = -\frac{13.6}{n_f^2} \] \[ -1.51 = -\frac{13.6}{n_f^2} \] \[ n_f^2 = \frac{13.6}{1.51} \approx 9 \] \[ n_f = 3 \] So, the electron jumps from the n=1 state to the n=3 state.
Part 2: Calculate 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 increase 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: Final Answer:
The angular momentum of the electron increases by 2(h/2\(\pi\)).