Find the value of wave number `(overset-v)` in terms of Rydberg's constant, when transition of electron takes place between two lvels of `He^(+)` ion whose sum is 4 and difference is 2.

Options

A. `(8R)/(9)`
B. `(32R)/(9)`
C. `(3R)/(4)`
D. none of these

Answer

To find the value of the wave number \( \overset{v}{\text{}} \) in terms of Rydberg's constant for the transition of an electron in the \( \text{He}^+ \) ion, we can follow these steps:

### Step 1: Define the Energy Levels
Let the two energy levels be \( n_1 \) and \( n_2 \). We are given two conditions:
1. The sum of the two levels:
\[ n_1 + n_2 = 4 \]
2. The difference of the two levels:
\[ n_1 - n_2 = 2 \]

### Step 2: Solve for \( n_1 \) and \( n_2 \)
We can solve these two equations simultaneously.

- Adding the two equations:
\[ (n_1 + n_2) + (n_1 - n_2) = 4 + 2 \]
This simplifies to:
\[ 2n_1 = 6 \implies n_1 = 3 \]

- Now, substituting \( n_1 = 3 \) back into the first equation:
\[ 3 + n_2 = 4 \implies n_2 = 1 \]

Thus, we have:
\[ n_1 = 3, \quad n_2 = 1 \]

### Step 3: Calculate the Wave Number
The wave number \( \overset{v}{\text{}} \) can be calculated using the formula:
\[ \overset{v}{\text{}} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant and \( Z \) is the atomic number. For \( \text{He}^+ \), \( Z = 2 \).

Substituting the values:
\[ \overset{v}{\text{}} = R(2^2) \left( \frac{1}{3^2} - \frac{1}{1^2} \right) \]
\[ = R(4) \left( \frac{1}{9} - 1 \right) \]
\[ = R(4) \left( \frac{1 - 9}{9} \right) \]
\[ = R(4) \left( \frac{-8}{9} \right) \]
\[ = -\frac{32R}{9} \]

Since wave number is typically expressed as a positive quantity, we can write:
\[ \overset{v}{\text{}} = \frac{32R}{9} \]

### Final Answer
Thus, the value of the wave number in terms of Rydberg's constant is:
\[ \overset{v}{\text{}} = \frac{32R}{9} \]

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