If the magnitude of energy of the electron in nth Bohr orbit is `E_n and J_n` is its angular momentum , then

Options

A. `E_npropJ_n^3`
B. `E_nprop1/J_n^(-3)`
C. `E_npropJ_n^2`
D. `E_nprop1/J_n^2`

Answer

To solve the problem, we need to analyze the relationships between the energy of an electron in the nth Bohr orbit and its angular momentum.

### Step-by-Step Solution:

1. **Understanding the Energy of the Electron:**
The energy \( E_n \) of an electron in the nth Bohr orbit is given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

2. **Understanding Angular Momentum:**
According to Bohr's model, the angular momentum \( J_n \) of the electron in the nth orbit is quantized and is given by:
\[ J_n = n \cdot \frac{h}{2\pi} \]
where \( h \) is Planck's constant.

3. **Relating Angular Momentum to Principal Quantum Number:**
From the angular momentum equation, we can express \( n \) in terms of \( J_n \):
\[ n = \frac{2\pi J_n}{h} \]

4. **Substituting \( n \) into the Energy Equation:**
Now, we substitute \( n \) back into the energy equation:
\[ E_n = -\frac{13.6 \, Z^2}{\left(\frac{2\pi J_n}{h}\right)^2} \]
Simplifying this gives:
\[ E_n = -\frac{13.6 \, Z^2 \cdot h^2}{(2\pi)^2 \cdot J_n^2} \]

5. **Identifying the Relationship:**
From the equation derived, we can see that the energy \( E_n \) is inversely proportional to the square of the angular momentum \( J_n \):
\[ E_n \propto -\frac{1}{J_n^2} \]

6. **Conclusion:**
Therefore, the correct option is that the energy \( E_n \) is inversely related to the square of the angular momentum \( J_n \).

### Final Answer:
The energy \( E_n \) is inversely related to the square of the angular momentum \( J_n \).