When the electron in hydrogen atom jumps from the second orbit to the first orbit, the wavelength of the radiation emitted is `lambda`. When the electron jumps from the third to the first orbit . The wavelenth of the radiation emitted is

Options

A. `(9)/(4)lambda`
B. `(4)/(9)lambda`
C. `(27)/(32)lambda`
D. `(32)/(27)lambda`

Answer

To solve the problem, we need to determine the wavelength of the radiation emitted when an electron in a hydrogen atom jumps from the third orbit (n=3) to the first orbit (n=1). We already know that when the electron jumps from the second orbit (n=2) to the first orbit (n=1), the wavelength of the emitted radiation is λ.

### Step-by-Step Solution:

1. **Energy of the emitted radiation**: The energy emitted when an electron transitions between two orbits in a hydrogen atom can be calculated using the formula:
\[ E = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number (which is 1 for hydrogen), and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy levels, respectively.

2. **First case (n=2 to n=1)**:
- Here, \( n_1 = 1 \) and \( n_2 = 2 \).
- Plugging in the values:
\[ E_1 = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]
- Thus, \( E_1 = \frac{3R}{4} \).

3. **Second case (n=3 to n=1)**:
- Here, \( n_1 = 1 \) and \( n_2 = 3 \).
- Plugging in the values:
\[ E_2 = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \cdot \frac{8}{9} \]
- Thus, \( E_2 = \frac{8R}{9} \).

4. **Relating the energies to wavelengths**:
- We know that energy is inversely proportional to wavelength:
\[ E \propto \frac{1}{\lambda} \]
- Therefore, we can write:
\[ \frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} \]
- Substituting the values we found:
\[ \frac{\frac{3R}{4}}{\frac{8R}{9}} = \frac{\lambda_2}{\lambda} \]
- Simplifying this gives:
\[ \frac{3}{4} \cdot \frac{9}{8} = \frac{\lambda_2}{\lambda} \]
\[ \frac{27}{32} = \frac{\lambda_2}{\lambda} \]

5. **Finding λ₂**:
- Rearranging gives:
\[ \lambda_2 = \frac{27}{32} \lambda \]

Thus, the wavelength of the radiation emitted when the electron jumps from the third orbit to the first orbit is:
\[ \lambda_2 = \frac{27}{32} \lambda \]

### Final Answer:
The wavelength of the radiation emitted when the electron jumps from the third to the first orbit is \( \frac{27}{32} \lambda \).

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