Question

The energy of an electron in Bohr's hydrogen atom is \(-3.4\,eV\). The angular momentum of the electron is

• A. \( \dfrac{2h}{\pi} \)
• B. \( \dfrac{h}{2\pi} \)
• C. \( \dfrac{h}{\pi} \)
• D. \( \dfrac{h}{4\pi} \)

Hint:

For Bohr atom: \[ E_n\propto \frac{1}{n^2} \] and \[ L_n=n\frac{h}{2\pi} \] Always determine \(n\) first from energy.

Answer 1

The Correct Option is C

Step 1: Recall the Bohr energy levels.
In a hydrogen atom the energy of the electron in level $n$ is \[ E_n = \frac{-13.6}{n^2}\,\text{eV} \]

Step 2: Match the given energy.
The electron has energy $-3.4$ eV. \[ -3.4 = \frac{-13.6}{n^2} \]
Step 3: Solve for n squared.
\[ n^2 = \frac{13.6}{3.4} = 4 \]
Step 4: Find n.
\[ n = 2 \]
Step 5: Use the angular momentum rule.
Bohr's rule says angular momentum is quantised. \[ L = n\frac{h}{2\pi} \]
Step 6: Put in n equals 2.
\[ L = 2 \times \frac{h}{2\pi} = \frac{h}{\pi} \] \[ \boxed{\dfrac{h}{\pi}} \]