The wavelength of radiation emitted is `lambda_(0)` when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of hydrogen atom, the wavelength of radiation emitted will be

Options

A. `(16//25)lambda_(0)`
B. `(20//27) lambda_(0)`
C. `(27//20) lambda_(0)`
D. `(25//16) lambda_(0)`

Answer

To solve the problem, we need to determine the wavelength of radiation emitted when an electron jumps from the fourth to the second orbit of a hydrogen atom. We will use the Rydberg formula for the Balmer series, which describes the wavelengths of the spectral lines of hydrogen.

### Step-by-Step Solution:

1. **Understand the Balmer Series**: The Balmer series describes transitions where electrons fall to the second energy level (n=2). The Rydberg formula for the wavelength of 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, \( n_1 \) is the lower energy level (2 in this case), and \( n_2 \) is the higher energy level.

2. **Case 1: Electron jumps from n=3 to n=2**: For the first case where the wavelength is \( \lambda_0 \):
- Here, \( n_1 = 2 \) and \( n_2 = 3 \).
- Applying the Rydberg formula:
\[ \frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
- Finding a common denominator (36):
\[ \frac{1}{\lambda_0} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]
- Thus, we have:
\[ \frac{1}{\lambda_0} = \frac{5R}{36} \]

3. **Case 2: Electron jumps from n=4 to n=2**: Now we need to find the wavelength when the electron jumps from the fourth orbit (n=4) to the second orbit (n=2):
- Here, \( n_1 = 2 \) and \( n_2 = 4 \).
- Applying the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
- Finding a common denominator (16):
\[ \frac{1}{\lambda} = R \left( \frac{4}{16} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \]

4. **Relate the two cases**: We can relate the two wavelengths:
- From the first case:
\[ \frac{1}{\lambda_0} = \frac{5R}{36} \]
- From the second case:
\[ \frac{1}{\lambda} = \frac{3R}{16} \]

5. **Find the ratio of the wavelengths**:
- Taking the ratio of \( \frac{1}{\lambda} \) and \( \frac{1}{\lambda_0} \):
\[ \frac{\lambda_0}{\lambda} = \frac{5R/36}{3R/16} = \frac{5 \cdot 16}{36 \cdot 3} = \frac{80}{108} = \frac{20}{27} \]
- Thus, we can express \( \lambda \) in terms of \( \lambda_0 \):
\[ \lambda = \frac{27}{20} \lambda_0 \]

6. **Final Result**: Therefore, the wavelength of radiation emitted when the electron jumps from the fourth to the second orbit of the hydrogen atom is:
\[ \lambda = \frac{20}{27} \lambda_0 \]