Question

The energy of an electron in an orbit in hydrogen atom is -3.4 eV. Its angular momentum in the orbit will be:

• A. \(\frac{3h}{2\pi}\)
• B. \(\frac{2h}{\pi}\)
• C. \(\frac{h}{\pi}\)
• D. \(\frac{h}{2\pi}\)

Hint:

Memorizing the energy levels for Hydrogen helps save time: $n=1$ is $-13.6$ eV, $n=2$ is $-3.4$ eV, and $n=3$ is $-1.51$ eV.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Bohr's model states that the energy of an electron in a hydrogen atom is quantized and specifically related to the principal quantum number $n$. Furthermore, the angular momentum of an electron is also quantized and is an integral multiple of $h/2\pi$.

Step 2: Key Formula or Approach:
1. Energy of $n^{th}$ orbit: \(E_n = -\frac{13.6}{n^2}\) eV
2. Bohr's quantization of angular momentum: \(L = \frac{nh}{2\pi}\)

Step 3: Detailed Explanation:
Given the energy \(E_n = -3.4\) eV: \[ -3.4 = -\frac{13.6}{n^2} \] \[ n^2 = \frac{13.6}{3.4} = 4 \] \[ n = 2 \] For the second orbit (\(n = 2\)), the angular momentum \(L\) is: \[ L = \frac{2h}{2\pi} = \frac{h}{\pi} \]

Step 4: Final Answer:
The angular momentum in the orbit is \(\frac{h}{\pi}\).