The wave number of energy emitted when electron jumbs from fourth orbit to second orbit in hydrogen in 20,497 `cm^(-1)`. The wave number of energy for the same transition in `He^(+)` is

Options

A. `5,099 cm^(-1)`
B. `20,497 cm^(-1)`
C. `40,994 cm^(-1)`
D. `81,988 cm^(-1)`

Answer

To find the wave number of energy emitted when an electron jumps from the fourth orbit to the second orbit in the helium ion \( He^+ \), we can use the Rydberg formula for the wave number of spectral lines:

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

where:
- \( \bar{\nu} \) is the wave number,
- \( Z \) is the atomic number,
- \( R \) is the Rydberg constant (approximately \( 109677 \, \text{cm}^{-1} \)),
- \( n_f \) is the final energy level,
- \( n_i \) is the initial energy level.

### Step-by-Step Solution:

1. **Identify the given values for Hydrogen:**
- The wave number for the transition from the fourth orbit to the second orbit in hydrogen is given as \( \bar{\nu}_{H} = 20497 \, \text{cm}^{-1} \).
- For hydrogen, \( Z = 1 \), \( n_f = 2 \), and \( n_i = 4 \).

2. **Apply the Rydberg formula for Hydrogen:**
\[ \bar{\nu}_{H} = Z^2 R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
Substituting the values for hydrogen:
\[ 20497 = 1^2 \cdot R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ 20497 = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
\[ 20497 = R \left( \frac{4 - 1}{16} \right) \]
\[ 20497 = R \cdot \frac{3}{16} \]

3. **Solve for Rydberg constant \( R \):**
\[ R = 20497 \cdot \frac{16}{3} \]
\[ R = 109318.67 \, \text{cm}^{-1} \]

4. **Now apply the Rydberg formula for \( He^+ \):**
- For \( He^+ \), \( Z = 2 \), \( n_f = 2 \), and \( n_i = 4 \).
\[ \bar{\nu}_{He^+} = 2^2 R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ \bar{\nu}_{He^+} = 4R \left( \frac{1}{4} - \frac{1}{16} \right) \]
\[ \bar{\nu}_{He^+} = 4R \cdot \frac{3}{16} \]
\[ \bar{\nu}_{He^+} = R \cdot \frac{12}{16} \]
\[ \bar{\nu}_{He^+} = R \cdot \frac{3}{4} \]

5. **Substituting the value of \( R \):**
\[ \bar{\nu}_{He^+} = 109318.67 \cdot \frac{3}{4} \]
\[ \bar{\nu}_{He^+} = 81989.00 \, \text{cm}^{-1} \]

### Final Answer:
The wave number of energy for the transition in \( He^+ \) is approximately \( 81989 \, \text{cm}^{-1} \).