What is the ratio of wavelength of radiations emitted when an electron in hydrogen atom jump from fourth orbit to second ornti and from third orbit to second orbit?

Options

A. `20 : 25`
B. `20 : 27`
C. `20 : 25`
D. `25 : 27`

Explanation

To find the ratio of the wavelengths of the radiations emitted when an electron in a hydrogen atom jumps from the fourth orbit to the second orbit and from the third orbit to the second orbit, we can use the formula for the wavelength of emitted radiation in a hydrogen atom.

### Step-by-Step Solution:

1. **Understanding the Formula**:
The formula for the wavelength (\(\lambda\)) of radiation emitted during an electron transition in a hydrogen atom is given by:
\[ \frac{1}{\lambda} = R \cdot (1/n_1^2 - 1/n_2^2) \]
where \(R\) is the Rydberg constant, \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.

2. **Transition from 4th Orbit to 2nd Orbit**:
- Here, \(n_1 = 2\) and \(n_2 = 4\).
- Plugging these values into the formula:
\[ \frac{1}{\lambda_1} = R \cdot \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R \cdot \left(\frac{1}{4} - \frac{1}{16}\right) \]
- Finding a common denominator (16):
\[ \frac{1}{\lambda_1} = R \cdot \left(\frac{4}{16} - \frac{1}{16}\right) = R \cdot \frac{3}{16} \]
- Therefore,
\[ \lambda_1 = \frac{16}{3R} \]

3. **Transition from 3rd Orbit to 2nd Orbit**:
- Here, \(n_1 = 2\) and \(n_2 = 3\).
- Plugging these values into the formula:
\[ \frac{1}{\lambda_2} = R \cdot \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R \cdot \left(\frac{1}{4} - \frac{1}{9}\right) \]
- Finding a common denominator (36):
\[ \frac{1}{\lambda_2} = R \cdot \left(\frac{9}{36} - \frac{4}{36}\right) = R \cdot \frac{5}{36} \]
- Therefore,
\[ \lambda_2 = \frac{36}{5R} \]

4. **Finding the Ratio of Wavelengths**:
- Now, we need to find the ratio \(\frac{\lambda_1}{\lambda_2}\):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{16}{3R}}{\frac{36}{5R}} = \frac{16}{3R} \cdot \frac{5R}{36} = \frac{16 \cdot 5}{3 \cdot 36} \]
- Simplifying:
\[ \frac{\lambda_1}{\lambda_2} = \frac{80}{108} = \frac{20}{27} \]

### Final Answer:
The ratio of the wavelengths of the radiations emitted when an electron jumps from the fourth orbit to the second orbit and from the third orbit to the second orbit is:
\[ \frac{20}{27} \]