When a transition of electron in `He^(+)` takes place from `n_(2)" to "n_(1)` then wave number in terms of Rydberg constant R will be
`("Given "n_(1)+n_(2)=4, n_(2)-n_(1)=2)`

Options

A. `(3R)/(4)`
B. `(8R)/(9)`
C. `(32R)/(9)`
D. (24R)/(9)`

Answer

To find the wave number for the transition of an electron in \( He^+ \) from \( n_2 \) to \( n_1 \), we can follow these steps:

### Step 1: Determine the values of \( n_1 \) and \( n_2 \)

We are given two equations:
1. \( n_1 + n_2 = 4 \)
2. \( n_2 - n_1 = 2 \)

From the first equation, we can express \( n_1 \) in terms of \( n_2 \):
\[ n_1 = 4 - n_2 \]

Now, substituting this expression into the second equation:
\[ n_2 - (4 - n_2) = 2 \]
Simplifying this gives:
\[ n_2 - 4 + n_2 = 2 \\ 2n_2 - 4 = 2 \\ 2n_2 = 6 \\ n_2 = 3 \]

Now substituting \( n_2 \) back into the equation for \( n_1 \):
\[ n_1 = 4 - n_2 = 4 - 3 = 1 \]

### Step 2: Use the Rydberg formula to find the wave number

The wave number \( \bar{\nu} \) (in terms of Rydberg constant \( R \)) for the transition can be calculated using the Rydberg formula:
\[ \bar{\nu} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For \( He^+ \), the atomic number \( Z \) is 2. Substituting \( n_1 = 1 \) and \( n_2 = 3 \):
\[ \bar{\nu} = R \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]

Calculating further:
\[ \bar{\nu} = R \cdot 4 \left( 1 - \frac{1}{9} \right) \\ = R \cdot 4 \left( \frac{9 - 1}{9} \right) \\ = R \cdot 4 \left( \frac{8}{9} \right) \\ = \frac{32R}{9} \]

### Conclusion

The wave number for the transition from \( n_2 \) to \( n_1 \) in \( He^+ \) is:
\[ \bar{\nu} = \frac{32R}{9} \]