Question

The energy of an electron in an orbit of the Bohr's atom is $-0.04 E_g$ where $E_g$ is the ground state energy. If $L$ is the angular momentum of the electron in this orbit and $h$ is the Planck's constant, then $\frac{2 \pi L{h}$ is _________} :}

• A. 4
• B. 6
• C. 5
• D. 2

Hint:

The term $\frac{2 \pi L}{h}$ is simply equal to the principal quantum number $n$ of the orbit in the Bohr model.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
According to the Bohr model, the energy of an electron in the $n^{th}$ orbit is inversely proportional to $n^2$, and the angular momentum is quantized.

Step 2: Key Formula or Approach:
1. Energy: \( E_n = \frac{E_g}{n^2} \)
2. Angular momentum: \( L = \frac{n h}{2 \pi} \)

Step 3: Detailed Explanation:
We are given that the energy in the orbit is:
\[ E_n = -0.04 E_g = -\frac{4}{100} E_g = -\frac{1}{25} E_g \]
Using the energy formula \( E_n = \frac{E_g}{n^2} \):
\[ \frac{1}{n^2} = \frac{1}{25} \implies n = 5 \]
The Bohr quantization of angular momentum states:
\[ L = \frac{n h}{2 \pi} \implies \frac{2 \pi L}{h} = n \]
Substituting \( n = 5 \):
\[ \frac{2 \pi L}{h} = 5 \]

Step 4: Final Answer:
The value of $\frac{2 \pi L}{h}$ is 5.