Question:
The energy of an electron in Bohr's hydrogen atom is \(-3.4\,eV\). The angular momentum of the electron is
The energy of an electron in Bohr's hydrogen atom is \(-3.4\,eV\). The angular momentum of the electron is
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For Bohr atom:
\[
E_n\propto \frac{1}{n^2}
\]
and
\[
L_n=n\frac{h}{2\pi}
\]
Always determine \(n\) first from energy.
Updated On: Jun 17, 2026
- \( \dfrac{2h}{\pi} \)
- \( \dfrac{h}{2\pi} \)
- \( \dfrac{h}{\pi} \)
- \( \dfrac{h}{4\pi} \)
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The Correct Option is C
Solution and Explanation
Concept:
Energy levels in Bohr atom:
\[
E_n=\frac{-13.6}{n^2}eV
\]
Angular momentum:
\[
L=n\frac{h}{2\pi}
\]
Step 1: Find principal quantum number. Given: \[ E_n=-3.4eV \] Using: \[ -3.4=\frac{-13.6}{n^2} \] \[ n^2=4 \] \[ n=2 \]
Step 2: Calculate angular momentum. \[ L=n\frac{h}{2\pi} \] \[ L=2\cdot\frac{h}{2\pi} \] \[ L=\frac{h}{\pi} \] Hence: \[ \boxed{\frac{h}{\pi}} \]
Step 1: Find principal quantum number. Given: \[ E_n=-3.4eV \] Using: \[ -3.4=\frac{-13.6}{n^2} \] \[ n^2=4 \] \[ n=2 \]
Step 2: Calculate angular momentum. \[ L=n\frac{h}{2\pi} \] \[ L=2\cdot\frac{h}{2\pi} \] \[ L=\frac{h}{\pi} \] Hence: \[ \boxed{\frac{h}{\pi}} \]
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