if the electron in hydrogen orbit jumps form third orbit to second orbit, the wavelength of the emitted radiation is given by

Options

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

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 (n1 = 3) to the second orbit (n2 = 2), we can use the Rydberg formula for hydrogen:

### Step-by-Step Solution:

1. **Understand the Rydberg Formula**:
The formula for the wavelength (λ) of the emitted radiation during an electron transition in a hydrogen atom is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the initial energy level, and \( n_2 \) is the final energy level.

2. **Identify the Values**:
For this problem:
- \( n_1 = 3 \) (initial orbit)
- \( n_2 = 2 \) (final orbit)

3. **Substitute the Values into the Formula**:
Plugging in the values into the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
This simplifies to:
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

4. **Calculate the Difference**:
To combine the fractions \( \frac{1}{4} \) and \( \frac{1}{9} \), we need a common denominator. The least common multiple of 4 and 9 is 36:
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
Therefore:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

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

6. **Solve for Wavelength (λ)**:
Rearranging to find \( \lambda \):
\[ \lambda = \frac{36}{5R} \]

7. **Conclusion**:
The wavelength of the emitted radiation when the electron jumps from the third orbit to the second orbit in a hydrogen atom is:
\[ \lambda = \frac{36}{5R} \]