If the electron in the hydrogen atom jumps from third orbit to second orbit the wavelength of the emitted radiation in term of Rydberg constant is

Options

A. `(6)/(5R)`
B. `(36)/(5R)`
C. `(64)/(7R)`
D. None of these

Explanation

To solve the problem of finding the wavelength of the emitted radiation when an electron in a hydrogen atom jumps from the third orbit to the second orbit, we can follow these steps:

### Step-by-Step Solution:

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

2. **Use the Rydberg Formula**:
- The formula to calculate the wavelength (λ) of the emitted radiation is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Here, \( R \) is the Rydberg constant, \( n_1 \) is the initial energy level, and \( n_2 \) is the final energy level.

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

4. **Calculate the Squares**:
- Calculate \( 3^2 \) and \( 2^2 \):
\[ 3^2 = 9 \quad \text{and} \quad 2^2 = 4 \]

5. **Calculate the Differences**:
- Now, substitute the squares back into the equation:
\[ \frac{1}{\lambda} = R \left( \frac{1}{9} - \frac{1}{4} \right) \]

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

7. **Perform the Subtraction**:
- Now, perform the subtraction:
\[ \frac{1}{\lambda} = R \left( \frac{4 - 9}{36} \right) = R \left( \frac{-5}{36} \right) \]

8. **Invert to Find Wavelength**:
- To find \( \lambda \), take the reciprocal:
\[ \lambda = \frac{36}{5R} \]

9. **Final Result**:
- Thus, the wavelength of the emitted radiation when the electron jumps from the third orbit to the second orbit is:
\[ \lambda = \frac{36}{5R} \]