The longest wavelength of ` He^(+)` in paschen series is "m", then shortest wavelenght of `Be^(+3)` in Pacchen series is( in terms of m):

Options

A. `(5)/(36)m`
B. `(64)/(7)m`
C. `(53)/(8)m`
D. `(7)/(64)m`

Explanation

To solve the problem, we need to find the shortest wavelength of \( Be^{+3} \) in the Paschen series, given that the longest wavelength of \( He^{+} \) in the same series is \( m \).

### Step-by-Step Solution:

1. **Understand the Paschen Series**:
The Paschen series involves transitions to the \( n=3 \) energy level. The wavelengths can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

2. **Calculate the Longest Wavelength for \( He^{+} \)**:
- For \( He^{+} \) (atomic number \( Z = 2 \)):
- The longest wavelength occurs for the transition \( n_1 = 3 \) and \( n_2 = 4 \):
\[ \frac{1}{\lambda_{He}} = R \cdot 2^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
- Simplifying this:
\[ \frac{1}{\lambda_{He}} = 4R \left( \frac{1}{9} - \frac{1}{16} \right) \]
- Finding a common denominator (144):
\[ \frac{1}{\lambda_{He}} = 4R \left( \frac{16 - 9}{144} \right) = 4R \left( \frac{7}{144} \right) = \frac{28R}{144} = \frac{7R}{36} \]
- Therefore, the longest wavelength \( \lambda_{He} = m \):
\[ m = \frac{36}{7R} \]

3. **Calculate the Shortest Wavelength for \( Be^{+3} \)**:
- For \( Be^{+3} \) (atomic number \( Z = 4 \)):
- The shortest wavelength occurs for the transition \( n_1 = 3 \) and \( n_2 = \infty \):
\[ \frac{1}{\lambda_{Be}} = R \cdot 4^2 \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) \]
- Since \( \frac{1}{\infty^2} = 0 \):
\[ \frac{1}{\lambda_{Be}} = 16R \cdot \frac{1}{9} = \frac{16R}{9} \]
- Therefore, the wavelength \( \lambda_{Be} \):
\[ \lambda_{Be} = \frac{9}{16R} \]

4. **Relate \( \lambda_{Be} \) to \( m \)**:
- We have \( m = \frac{36}{7R} \) and \( \lambda_{Be} = \frac{9}{16R} \).
- To express \( \lambda_{Be} \) in terms of \( m \):
\[ \lambda_{Be} = \frac{9}{16R} = \frac{9}{16} \cdot \frac{36}{7m} = \frac{324}{112m} = \frac{81}{28m} \]

5. **Final Answer**:
The shortest wavelength of \( Be^{+3} \) in terms of \( m \) is:
\[ \lambda_{Be} = \frac{81}{28} m \]