In a mixture of `He^(o+)` gas H atom and `He^(o+)` ions Are excited to three respective first excited subsepuenly , H atom transfers its total excitation energy to `He^(o+)`ions by collision .Assuming that Bohr model of an atom is applicable , answer the following question
The ratio of teh potential energy of the `n = 2` electron for H atom to the of `He^(o+)` ion is

Options

A. `1//4`
B. `1//2`
C. 4
D. 3

Explanation

To solve the problem, we need to find the ratio of the potential energy of the \( n = 2 \) electron for a hydrogen atom (\( H \)) to that of a helium ion (\( He^{+} \)). We will use the Bohr model of the atom to derive the necessary formulas.

### Step-by-Step Solution:

1. **Understand the Potential Energy Formula**:
The potential energy (\( PE \)) of an electron in a hydrogen-like atom can be expressed in terms of its total energy (\( E \)). The relationship is given by:
\[ PE = 2 \times E \]

2. **Total Energy for Hydrogen and Helium Ion**:
The total energy (\( E \)) of an electron in a hydrogen-like atom is given by the formula:
\[ E = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

3. **Calculate Total Energy for Hydrogen Atom**:
For hydrogen (\( H \)), \( Z = 1 \) and \( n = 2 \):
\[ E_H = -\frac{13.6 \times 1^2}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]

4. **Calculate Potential Energy for Hydrogen Atom**:
Using the relationship \( PE = 2 \times E \):
\[ PE_H = 2 \times (-3.4) = -6.8 \, \text{eV} \]

5. **Calculate Total Energy for Helium Ion**:
For \( He^{+} \), \( Z = 2 \) and \( n = 2 \):
\[ E_{He^+} = -\frac{13.6 \times 2^2}{2^2} = -\frac{13.6 \times 4}{4} = -13.6 \, \text{eV} \]

6. **Calculate Potential Energy for Helium Ion**:
Using the relationship \( PE = 2 \times E \):
\[ PE_{He^+} = 2 \times (-13.6) = -27.2 \, \text{eV} \]

7. **Calculate the Ratio of Potential Energies**:
Now, we can find the ratio of the potential energy of hydrogen to that of helium ion:
\[ \text{Ratio} = \frac{PE_H}{PE_{He^+}} = \frac{-6.8}{-27.2} = \frac{6.8}{27.2} = \frac{1}{4} \]

### Final Answer:
The ratio of the potential energy of the \( n = 2 \) electron for the hydrogen atom to that of the helium ion is:
\[ \frac{1}{4} \]