Question

Angular momentum of an electron in a hydrogen atom is \( \dfrac{3h}{\pi} \). Then the energy of the electron is ____ eV.

• A. \(-1.51\)
• B. \(-0.85\)
• C. \(-0.38\)
• D. \(-0.28\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
According to Bohr's model of the hydrogen atom, angular momentum is quantized as an integral multiple of \(\frac{h}{2\pi}\). We first determine the principal quantum number \(n\) and then find the corresponding energy level.
Step 2: Key Formula or Approach:
1. Bohr's quantization condition: \(L = n \left( \frac{h}{2\pi} \right)\).
2. Energy of \(n\)-th orbit: \(E_n = -\frac{13.6}{n^2} \text{ eV}\).
Step 3: Detailed Explanation:
Given angular momentum \(L = \frac{3h}{\pi}\).
Using the quantization condition:
\[ \frac{nh}{2\pi} = \frac{3h}{\pi} \]
\[ \frac{n}{2} = 3 \implies n = 6 \].
The electron is in the 6th orbit.
Now, calculate the energy:
\[ E_6 = -\frac{13.6}{6^2} = -\frac{13.6}{36} \approx -0.377 \text{ eV} \].
Rounding off, we get approximately -0.38 eV.
Step 4: Final Answer:
The energy of the electron is -0.38 eV.