If the electron in a hydrogen atom jumps from an orbit with level `n_(2)=3` to orbit with level `n_(1)=2`, the emitted radiation has a wavelength given by .

Options

A. `lambda =(36)/(5R)`
B. `lambda=(5R)/(36)`
C. `lambda=(6)/( R)`
D. `lambda=( R )/(6)`.

Explanation

To find the wavelength of the emitted radiation when an electron in a hydrogen atom jumps from an orbit with level \( n_2 = 3 \) to an orbit with level \( n_1 = 2 \), we can use the Rydberg formula for hydrogen:

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

Where:
- \( \lambda \) is the wavelength of the emitted radiation,
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the lower energy level (2 in this case),
- \( n_2 \) is the higher energy level (3 in this case).

### Step-by-step Solution:

1. **Identify the values of \( n_1 \) and \( n_2 \)**:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

2. **Substitute the values into the Rydberg formula**:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

3. **Calculate \( \frac{1}{2^2} \) and \( \frac{1}{3^2} \)**:
- \( \frac{1}{2^2} = \frac{1}{4} \)
- \( \frac{1}{3^2} = \frac{1}{9} \)

4. **Find a common denominator to subtract the fractions**:
- The common denominator of 4 and 9 is 36.
- Convert \( \frac{1}{4} \) and \( \frac{1}{9} \) to have the same denominator:
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]

5. **Subtract the fractions**:
\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

6. **Substitute back into the Rydberg formula**:
\[ \frac{1}{\lambda} = R \left( \frac{5}{36} \right) \]

7. **Rearranging to find \( \lambda \)**:
\[ \lambda = \frac{36}{5R} \]

8. **Final Expression**:
- The wavelength of the emitted radiation when the electron jumps from \( n_2 = 3 \) to \( n_1 = 2 \) is:
\[ \lambda = \frac{36}{5R} \]