The wavelength of the radiation emitted is `lambda_0` when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be `20/x lambda_0` . The value of x is_____.

Answer

To solve the problem, we need to find the value of \( x \) based on the given information about the transitions of an electron in a hydrogen atom.

### Step-by-Step Solution:

1. **Identify the transitions**:
- The first transition is from the second excited state (n=3) to the first excited state (n=2).
- The second transition is from the third excited state (n=4) to the second orbit (n=2).

2. **Calculate the wavelength for the first transition**:
- The formula for the wavelength emitted during a transition in a hydrogen atom is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For the transition from n=3 to n=2:
\[ \frac{1}{\lambda_0} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{9 - 4}{36} \right) = R_H \left( \frac{5}{36} \right) \]
- Thus, we can express \( \lambda_0 \):
\[ \lambda_0 = \frac{36}{5R_H} \]

3. **Calculate the wavelength for the second transition**:
- For the transition from n=4 to n=2:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{4 - 1}{16} \right) = R_H \left( \frac{3}{16} \right) \]
- Thus, we can express \( \lambda \):
\[ \lambda = \frac{16}{3R_H} \]

4. **Relate the two wavelengths**:
- According to the problem, the wavelength for the second transition is given as:
\[ \lambda = \frac{20}{x} \lambda_0 \]
- Substituting the expressions for \( \lambda \) and \( \lambda_0 \):
\[ \frac{16}{3R_H} = \frac{20}{x} \cdot \frac{36}{5R_H} \]
- Cancel \( R_H \) from both sides:
\[ \frac{16}{3} = \frac{20 \cdot 36}{5x} \]
- Simplifying the right side:
\[ \frac{16}{3} = \frac{720}{5x} \]
\[ \frac{16}{3} = \frac{144}{x} \]

5. **Cross-multiply to solve for \( x \)**:
- Cross-multiplying gives:
\[ 16x = 432 \]
- Dividing both sides by 16:
\[ x = \frac{432}{16} = 27 \]

### Final Answer:
The value of \( x \) is **27**.