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 solve the problem, we need to find the wave number when an electron transitions from level \( n_2 \) to \( n_1 \) in the helium ion \( He^+ \). The given conditions are:

1. \( n_1 + n_2 = 4 \)
2. \( n_2 - n_1 = 2 \)

### Step 1: Solve for \( n_1 \) and \( n_2 \)

We have two equations:

1. \( n_1 + n_2 = 4 \) (Equation 1)
2. \( n_2 - n_1 = 2 \) (Equation 2)

To find \( n_1 \) and \( n_2 \), we can add these two equations together:

\[ (n_1 + n_2) + (n_2 - n_1) = 4 + 2 \]

This simplifies to:

\[ 2n_2 = 6 \]

Dividing both sides by 2 gives:

\[ n_2 = 3 \]

Now, substitute \( n_2 \) back into Equation 1 to find \( n_1 \):

\[ n_1 + 3 = 4 \]

So,

\[ n_1 = 1 \]

### Step 2: Use the wave number formula

The wave number \( \bar{\nu} \) for a transition in a hydrogen-like atom is given by the formula:

\[ \bar{\nu} = R z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R \) is the Rydberg constant,
- \( z \) is the atomic number (for \( He^+ \), \( z = 2 \)),
- \( n_1 = 1 \) and \( n_2 = 3 \).

### Step 3: Substitute the values into the formula

Now, substituting \( n_1 \), \( n_2 \), and \( z \) into the wave number formula:

\[ \bar{\nu} = R (2^2) \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]

Calculating \( z^2 \):

\[ 2^2 = 4 \]

Now substituting this into the equation:

\[ \bar{\nu} = R \cdot 4 \left( 1 - \frac{1}{9} \right) \]

Calculating \( 1 - \frac{1}{9} \):

\[ 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \]

Now substituting this back into the wave number equation:

\[ \bar{\nu} = R \cdot 4 \cdot \frac{8}{9} \]

Calculating this gives:

\[ \bar{\nu} = \frac{32R}{9} \]

### Final Answer

The wave number in terms of the Rydberg constant \( R \) is:

\[ \bar{\nu} = \frac{32R}{9} \]