If an electron in hydrogen atom jumps from third orbit to second orbit, the Wavelength of the emitted radiation is given by (c is speed of light )

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the Initial and Final States**:
- The electron is moving from the third orbit (n2 = 3) to the second orbit (n1 = 2).

2. **Use the Rydberg Formula**:
The formula for the wavelength (λ) of the emitted radiation is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant.

3. **Substitute the Values**:
- Here, \( n_1 = 2 \) and \( n_2 = 3 \).
- Substitute these values into the formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

4. **Calculate the Squares**:
- Calculate \( 2^2 = 4 \) and \( 3^2 = 9 \).
- Now plug these values into the equation:
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

5. **Find a Common Denominator**:
- The common denominator for 4 and 9 is 36.
- Rewrite the fractions:
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
- Therefore:
\[ \frac{1}{\lambda} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]

6. **Rearrange to Find λ**:
- Now, we can rearrange the equation to find λ:
\[ \lambda = \frac{36}{5R} \]

7. **Final Result**:
- Thus, the wavelength of the emitted radiation is:
\[ \lambda = \frac{36}{5R} \]

### Conclusion:
The correct answer is \( \lambda = \frac{36R}{5} \).