Question

A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of: (Given, Planck’s constant = \(6.6 × 10^{–34}\) Js).

• A. \(2.10 × 10^{–34}\) Js
• B. \(1.05 × 10^{–34}\) Js
• C. \(3.15 × 10^{–34}\) Js
• D. \(4.2 × 10^{–34}\) Js

Answer 1

The Correct Option is B

To solve this question, we need to understand how the energy absorbed by a hydrogen atom affects the angular momentum of its electron.

When a hydrogen atom in its ground state absorbs energy, its electron transitions to a higher energy level. The energy absorbed by the hydrogen atom to transition from one state to another corresponds to specific energy differences between these states.

The ground state energy level of hydrogen (n=1) absorbing 10.2 eV indicates it is moving to the first excited state (n=2). The equation for the energy difference between two levels in a hydrogen atom is given by the Rydberg formula:

E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \, \text{eV}

In this case, n_1 = 1 and n_2 = 2.

Substituting these values:

E = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \cdot \frac{3}{4} = 10.2 \, \text{eV}

This confirms the transition from n=1 to n=2.

The angular momentum of an electron in a hydrogen atom is quantized and given by:

L = n \cdot \hbar, where n is the principal quantum number, and \hbar = \frac{h}{2\pi}

When transitioning from n=1 to n=2, the increase in angular momentum is:

\Delta L = \hbar(2 - 1) = \hbar

Substitute \hbar :

\hbar = \frac{h}{2\pi} = \frac{6.6 \times 10^{-34} \, \text{Js}}{2\pi}

Calculating, we have:

\hbar \approx \frac{6.6 \times 10^{-34}}{6.28} = 1.05 \times 10^{-34} \, \text{Js}

Thus, the angular momentum increases by 1.05 \times 10^{-34} \, \text{Js}.

Therefore, the correct answer is:

1.05 \times 10^{-34} Js