Check the corretness of the following statement about the Bohr model of hydrogen atom:
(i) The acceleration of the electron in `n = 2` orbit is more than that in `n = 1` orbit.
(ii) The angular momentum of the electron in `n = 2` orbit is more than that in `n = 1` orbit.
(iii) The `KE` of the electron in `n = 2` orbit is more than that in `n = 1` orbit.

Options

A. All the statements are correct.
B. Only (i) and (ii) are correct.
C. Only (ii) and (iii) are correct.
D. Only (iii) and (i) are correct.

Explanation

To solve the question regarding the correctness of the statements about the Bohr model of the hydrogen atom, we will analyze each statement one by one.

### Step-by-Step Solution:

1. **Analyze Statement (i)**: "The acceleration of the electron in `n = 2` orbit is more than that in `n = 1` orbit."
- The centripetal acceleration \( a \) of the electron in a circular orbit is given by the formula:
\[ a = \frac{v^2}{r} \]
- According to the Bohr model, the radius \( r \) of the orbit is given by:
\[ r_n = n^2 \cdot r_1 \]
where \( r_1 \) is the radius of the first orbit.
- As \( n \) increases, the radius \( r \) increases, which means that the acceleration decreases because the electron is moving in a larger circle.
- Therefore, the acceleration in the \( n = 2 \) orbit is **less** than that in the \( n = 1 \) orbit.
- **Conclusion**: This statement is **false**.

2. **Analyze Statement (ii)**: "The angular momentum of the electron in `n = 2` orbit is more than that in `n = 1` orbit."
- The angular momentum \( L \) of the electron in the Bohr model is given by:
\[ L = n \cdot \frac{h}{2\pi} \]
- Since \( n \) is greater for the \( n = 2 \) orbit than for the \( n = 1 \) orbit, the angular momentum in the \( n = 2 \) orbit will be greater.
- **Conclusion**: This statement is **true**.

3. **Analyze Statement (iii)**: "The `KE` of the electron in `n = 2` orbit is more than that in `n = 1` orbit."
- The kinetic energy \( KE \) of the electron in the Bohr model is given by:
\[ KE = \frac{1}{2} mv^2 \]
- The kinetic energy is also related to the principal quantum number \( n \) as follows:
\[ KE \propto \frac{1}{n^2} \]
- Therefore, as \( n \) increases, the kinetic energy decreases. Thus, the kinetic energy in the \( n = 2 \) orbit is **less** than that in the \( n = 1 \) orbit.
- **Conclusion**: This statement is **false**.

### Final Summary:
- Statement (i): **False**
- Statement (ii): **True**
- Statement (iii): **False**

Thus, the correct statements are only (ii).