In a Hydrogen, electron jumps from fourth orbit to second orbit. The wave number of the radiations emitted by electron is

Options

A. `R/16`
B. `(3R)/16`
C. `(5R)/16`
D. `(7R)/16`

Answer

To find the wave number of the radiation emitted when an electron in a hydrogen atom jumps from the fourth orbit to the second orbit, we can use the formula for wave number:

\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( \lambda \) is the wavelength of the emitted radiation,
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level (final orbit, which is 2),
- \( n_2 \) is the higher energy level (initial orbit, which is 4).

### Step-by-step Solution:

1. **Identify the values**:
- For hydrogen, \( Z = 1 \).
- The electron jumps from \( n_2 = 4 \) to \( n_1 = 2 \).

2. **Substitute the values into the formula**:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) \]

3. **Calculate \( \frac{1}{4} - \frac{1}{16} \)**:
- Convert \( \frac{1}{4} \) to a fraction with a denominator of 16:
\[ \frac{1}{4} = \frac{4}{16} \]
- Now subtract:
\[ \frac{4}{16} - \frac{1}{16} = \frac{3}{16} \]

4. **Substitute back into the wave number formula**:
\[ \frac{1}{\lambda} = R \cdot \frac{3}{16} \]

5. **Final expression for the wave number**:
\[ \frac{1}{\lambda} = \frac{3R}{16} \]

Thus, the wave number of the radiation emitted when the electron jumps from the fourth orbit to the second orbit in a hydrogen atom is:

\[ \frac{3R}{16} \]