Question

Angular momentum of the electron in a hydrogen atom is $\frac{3h}{2\pi}$, then find the energy of the electron in the orbit :-

• A. $-13.6 \text{ eV}$
• B. $-3.4 \text{ eV}$
• C. $-1.51 \text{ eV}$
• D. $-0.85 \text{ eV}$

Answer 1

The Correct Option is C

Step 1: Use Bohr's quantization rule to find the orbit number $n$.
$L = \frac{nh}{2\pi} = \frac{3h}{2\pi} \implies n = 3$.

Step 2: Use the energy formula for hydrogen atom.
$E_n = -\frac{13.6}{n^2} \text{ eV}$.

Step 3: Calculate energy for $n=3$.
$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \text{ eV}$.

Final Answer: Option (3).