In a hydrogen atom, the transition takes place from n = 3 to n = 2 . If Rydberg constant is `1.097xx10^(7)m^(-1)` , the wavelength of the emitted radiation is

Options

A. `(36)/(5R_H)`
B. `(5R_H)/36`
C. `(3)/(4R_H)`
D. `(4)/(3R_H)`

Answer

To find the wavelength of the emitted radiation when an electron in a hydrogen atom transitions from n = 3 to n = 2, we can use the Rydberg formula for hydrogen:

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

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

### Step-by-Step Solution:

1. **Identify the values**:
- \(n_1 = 2\)
- \(n_2 = 3\)
- \(R_H = 1.097 \times 10^7 \, m^{-1}\)

2. **Substitute the values into the Rydberg formula**:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

3. **Calculate \(n_1^2\) and \(n_2^2\)**:
- \(n_1^2 = 2^2 = 4\)
- \(n_2^2 = 3^2 = 9\)

4. **Substitute \(n_1^2\) and \(n_2^2\) into the formula**:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) \]

5. **Calculate the difference inside the parentheses**:
- Find a common denominator (36):
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

6. **Substitute back into the equation**:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{5}{36} \]

7. **Calculate \(\frac{1}{\lambda}\)**:
\[ \frac{1}{\lambda} = \frac{1.097 \times 10^7 \times 5}{36} = \frac{5.485 \times 10^7}{36} \approx 1.524 \times 10^6 \, m^{-1} \]

8. **Calculate \(\lambda\)**:
\[ \lambda = \frac{1}{1.524 \times 10^6} \approx 6.55 \times 10^{-7} \, m \]

9. **Convert to nanometers (1 m = \(10^9\) nm)**:
\[ \lambda \approx 655 \, nm \]

### Final Answer:
The wavelength of the emitted radiation is approximately **655 nm**.