The angular momentum of an electron in a Bohr's orbit of `He^+` is `3.1652 xx 10^-34 kg-m^2//sec`. What is the wave number in terms of Rydberg constant `(R )` of the spectral line emitted when an electron falls this level to the first excited state.
`["Use" h = 6.626 xx 10^-34 Js]`.

Options

A. `3 R`
B. `(5 R)/(9)`
C. `(3 R)/(4)`
D. `(8 R)/(9)`

Explanation

To solve the problem, we need to find the wave number of the spectral line emitted when an electron transitions from the third energy level (n=3) to the first excited state (n=2) in the helium ion (He⁺). We will use the Rydberg formula for this calculation.

### Step-by-Step Solution:

1. **Understand the Transition**:
- The electron is transitioning from n=3 to n=2.
- The wave number (1/λ) can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Here, R is the Rydberg constant, Z is the atomic number (for He⁺, Z=2), \( n_1 \) is the lower energy level (2), and \( n_2 \) is the higher energy level (3).

2. **Substituting Values**:
- Substitute the values into the Rydberg formula:
\[ \frac{1}{\lambda} = R \cdot (2^2) \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
- This simplifies to:
\[ \frac{1}{\lambda} = R \cdot 4 \left( \frac{1}{4} - \frac{1}{9} \right) \]

3. **Calculating the Difference**:
- Calculate \( \frac{1}{4} - \frac{1}{9} \):
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]

4. **Final Calculation**:
- Substitute this back into the equation:
\[ \frac{1}{\lambda} = R \cdot 4 \cdot \frac{5}{36} \]
\[ \frac{1}{\lambda} = \frac{20R}{36} = \frac{5R}{9} \]

5. **Conclusion**:
- The wave number in terms of the Rydberg constant \( R \) is:
\[ \frac{1}{\lambda} = \frac{5R}{9} \]

### Final Answer:
The wave number of the spectral line emitted when an electron falls from the third level to the first excited state in He⁺ is \( \frac{5R}{9} \).