the wavelength of radiation emitted is `lambda_0` when an electron jumps. From the third to second orbit of hydrogen atom. For the electron jumping from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be

Options

A. `(16//25)lambda_0`
B. `(20//27)lambda_0`
C. `(27//20)lambda_0`
D. `(25//16)lambda_0`

Explanation

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

1. **Identify the initial and final states**:
- For the first case (from the third to the second orbit), we have:
- \( n_{initial} = 3 \)
- \( n_{final} = 2 \)

- For the second case (from the fourth to the second orbit), we have:
- \( n_{initial} = 4 \)
- \( n_{final} = 2 \)

2. **Write the Rydberg formula**:
The Rydberg formula for the wavelength \( \lambda \) of emitted radiation is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \]
where \( R \) is the Rydberg constant.

3. **Calculate for the first case** (from \( n = 3 \) to \( n = 2 \)):
\[ \frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
To combine the fractions:
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
Thus,
\[ \frac{1}{\lambda_0} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]

4. **Calculate for the second case** (from \( n = 4 \) to \( n = 2 \)):
\[ \frac{1}{\lambda'} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
To combine the fractions:
\[ \frac{1}{4} = \frac{4}{16}, \quad \frac{1}{16} = \frac{1}{16} \]
Thus,
\[ \frac{1}{\lambda'} = R \left( \frac{4 - 1}{16} \right) = R \left( \frac{3}{16} \right) \]

5. **Relate the two wavelengths**:
Now we have:
\[ \frac{1}{\lambda_0} = R \left( \frac{5}{36} \right) \quad \text{and} \quad \frac{1}{\lambda'} = R \left( \frac{3}{16} \right) \]

6. **Divide the two equations**:
\[ \frac{\lambda'}{\lambda_0} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{3}{16} \times \frac{36}{5} = \frac{108}{80} = \frac{27}{20} \]

7. **Rearranging gives**:
\[ \lambda' = \lambda_0 \times \frac{27}{20} \]

8. **Expressing in terms of \( \lambda_0 \)**:
\[ \lambda' = \frac{27}{20} \lambda_0 \]

Thus, the wavelength of radiation emitted when the electron jumps from the fourth to the second orbit is \( \frac{27}{20} \lambda_0 \).