Consider a system containing a negatively charge point `(pi, m_(pi) = 273m_(e))` orbital around a stationary nucleus of atomic number Z .The total energy `(E_(n))` of ion is half of its potential energy `(PE_(n))` in nth stationary state .The motion of the point can be assumed to be in a uniform circular motion with centripetal force given by the force of attraction between the positive nucleus and the point .Assume that point revolves only in the stationary state defined by the quantization of its angular momentum about the nucleus as Bohr's model
The potential energy `(PE_(n))` of ion follows:

Options

A. `PE_(n) prop m_(pi) ((n^(2))/(Z))`
B. `PE_(n) prop m_(pi) ((Z^(2))/(n^(2)))`
C. `PE_(n) prop (1)/(m_(pi)) ((n^(2))/(Z^(2)))`
D. `PE_(n) prop (1)/(m_(pi)) ((Z^(2))/(n^(2)))`

Explanation

To solve the problem, we need to derive the potential energy \( PE_n \) of the ion in the nth stationary state, based on the information provided in the question and the video transcript.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a negatively charged point particle (let's call it \( \pi \)) with mass \( m_{\pi} = 273 m_e \) (where \( m_e \) is the mass of the electron) orbiting around a stationary nucleus with atomic number \( Z \).
- The charge of the nucleus is \( +Ze \), and the charge of the point particle \( \pi \) is \( -e \).

2. **Centripetal Force and Electrostatic Force**:
- The centripetal force required for the circular motion of the particle is provided by the electrostatic force of attraction between the nucleus and the point charge.
- The centripetal force \( F_c \) can be expressed as:
\[ F_c = \frac{m_{\pi} v^2}{R} \]
- The electrostatic force \( F_e \) is given by Coulomb's law:
\[ F_e = \frac{K \cdot Ze \cdot e}{R^2} \]
- Setting these two forces equal gives:
\[ \frac{m_{\pi} v^2}{R} = \frac{K \cdot Ze^2}{R^2} \]

3. **Solving for Radius \( R \)**:
- Rearranging the equation to find \( R \):
\[ R = \frac{K \cdot Ze^2}{m_{\pi} v^2} \]

4. **Quantization of Angular Momentum**:
- According to Bohr's model, the angular momentum \( L \) is quantized:
\[ L = m_{\pi} v R = \frac{n h}{2\pi} \]
- Substituting \( R \) from the previous step into this equation gives:
\[ m_{\pi} v \left( \frac{K \cdot Ze^2}{m_{\pi} v^2} \right) = \frac{n h}{2\pi} \]
- Simplifying this leads to:
\[ K \cdot Ze^2 = \frac{n h v}{2\pi} \]

5. **Finding Velocity \( v \)**:
- Rearranging gives us an expression for \( v \):
\[ v = \frac{2\pi K Ze^2}{n h} \]

6. **Substituting \( v \) back to find \( R \)**:
- Substitute \( v \) into the expression for \( R \):
\[ R = \frac{K \cdot Ze^2}{m_{\pi} \left( \frac{2\pi K Ze^2}{n h} \right)^2} \]
- This simplifies to:
\[ R = \frac{n^2 h^2}{4\pi^2 m_{\pi} Z e^2} \]

7. **Calculating Potential Energy \( PE_n \)**:
- The potential energy \( PE_n \) is given by:
\[ PE_n = -\frac{K \cdot Ze^2}{R} \]
- Substituting \( R \) gives:
\[ PE_n = -\frac{K \cdot Ze^2 \cdot 4\pi^2 m_{\pi} Z e^2}{n^2 h^2} \]
- This simplifies to:
\[ PE_n = -\frac{4\pi^2 K Z^2 e^4 m_{\pi}}{n^2 h^2} \]

### Final Expression:
Thus, the potential energy \( PE_n \) of the ion in the nth stationary state can be expressed as:
\[ PE_n = -\frac{4\pi^2 K Z^2 e^4 m_{\pi}}{n^2 h^2} \]