When an electron jumps from the orbit n=2 to n=4, then wavelength of the radiations absorbed will be where, (where R is Rydberg's constant)

Options

A. `(16)/(3R)`
B. `(16)/(5R)`
C. `(5R)/(16)`
D. `(3R)/(16)`

Explanation

To solve the problem of finding the wavelength of radiation absorbed when an electron jumps from the orbit n=2 to n=4, we can use the Rydberg formula for hydrogen-like atoms. The formula is given by:

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

Where:
- \(\lambda\) is the wavelength of the emitted or absorbed radiation,
- \(R\) is the Rydberg constant,
- \(n_1\) is the lower energy level (initial state),
- \(n_2\) is the higher energy level (final state).

### Step-by-Step Solution:

1. **Identify the initial and final states**:
- The electron is jumping from \(n_1 = 2\) (initial state) to \(n_2 = 4\) (final state).

2. **Substitute the values into the Rydberg formula**:
- Plugging \(n_1\) and \(n_2\) into the formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

3. **Calculate the squares**:
- Calculate \(2^2\) and \(4^2\):
\[ 2^2 = 4 \quad \text{and} \quad 4^2 = 16 \]

4. **Substitute the squares back into the equation**:
- Now substitute these values into the equation:
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) \]

5. **Find a common denominator**:
- The common denominator for \(4\) and \(16\) is \(16\):
\[ \frac{1}{4} = \frac{4}{16} \]
- Therefore:
\[ \frac{1}{\lambda} = R \left( \frac{4}{16} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \]

6. **Simplify the equation**:
- Now we have:
\[ \frac{1}{\lambda} = \frac{3R}{16} \]

7. **Find the wavelength**:
- To find \(\lambda\), take the reciprocal:
\[ \lambda = \frac{16}{3R} \]

### Final Answer:
The wavelength of the radiation absorbed when the electron jumps from \(n=2\) to \(n=4\) is:
\[ \lambda = \frac{16}{3R} \]