In Bohr's model of hydrogen atom ,

Options

A. the radius of nth orbit is proportional to `n^(2)`
B. the total energy of electron in nth orbit is proportional to `n`
C. the amgular momentum of the electron in an orbit is an integral multiple of `h//2 pi`
D. the magnitude of the potential energy of an electron in any orbit is greater than its kinetic energy

Explanation

To solve the question regarding Bohr's model of the hydrogen atom, we will analyze each statement provided in the options and determine their correctness based on the principles of Bohr's theory.

### Step-by-Step Solution:

1. **Radius of nth Orbit:**
- According to Bohr's model, the radius of the nth orbit (Rn) for a hydrogen atom (Z = 1) is given by:
\[ R_n = \frac{R_0 n^2}{Z} \]
where \( R_0 \) is the ground state radius (approximately 0.529 Å).
- For hydrogen (Z = 1), this simplifies to:
\[ R_n = R_0 n^2 \]
- Therefore, the radius of the nth orbit is indeed proportional to \( n^2 \).
- **Conclusion:** This statement is **correct**.

2. **Total Energy of Electron in nth Orbit:**
- The total energy (E) of an electron in the nth orbit is given by:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
- For hydrogen (Z = 1), this becomes:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
- This indicates that the total energy is inversely proportional to \( n^2 \), not \( n \).
- **Conclusion:** This statement is **incorrect**.

3. **Angular Momentum of the Electron:**
- Bohr's model states that the angular momentum (L) of an electron in an orbit is quantized and given by:
\[ L = n \frac{h}{2\pi} \]
- Here, \( n \) is the principal quantum number, and \( h \) is Planck's constant.
- This means that the angular momentum is indeed an integral multiple of \( \frac{h}{2\pi} \).
- **Conclusion:** This statement is **correct**.

4. **Potential Energy vs. Kinetic Energy:**
- In Bohr's model, the relationship between kinetic energy (KE) and potential energy (PE) is given by:
\[ KE = -\frac{1}{2} E \]
and
\[ PE = 2 \cdot KE \]
- Therefore, the magnitude of potential energy is greater than that of kinetic energy, as:
\[ |PE| = 2 |KE| \]
- **Conclusion:** This statement is **correct**.

### Final Answer:
- The correct options are:
- A: Correct (Radius of nth orbit is proportional to \( n^2 \))
- B: Incorrect (Total energy is inversely proportional to \( n^2 \))
- C: Correct (Angular momentum is an integral multiple of \( \frac{h}{2\pi} \))
- D: Correct (Magnitude of potential energy is greater than kinetic energy)