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)`

Explanation

To solve the problem, we need to calculate the wavelength of radiation emitted when an electron jumps from the fourth orbit (n=4) to the second orbit (n=2) of a hydrogen atom, given that the wavelength for the transition from the third orbit (n=3) to the second orbit (n=2) is \( \lambda_0 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that when an electron transitions between energy levels in a hydrogen atom, it emits or absorbs radiation, which can be described by the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

2. **Calculate Wavelength for n=3 to n=2**:
For the transition from n=3 to n=2, we can write:
\[ \frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Simplifying this:
\[ \frac{1}{\lambda_0} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]

3. **Calculate Wavelength for n=4 to n=2**:
Now, for the transition from n=4 to n=2, we write:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
Simplifying this:
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4 - 1}{16} \right) = R \left( \frac{3}{16} \right) \]

4. **Finding the Ratio of Wavelengths**:
We can find the ratio of the two wavelengths:
\[ \frac{\lambda}{\lambda_0} = \frac{\frac{1}{R \left( \frac{3}{16} \right)}}{\frac{1}{R \left( \frac{5}{36} \right)}} = \frac{5}{3} \cdot \frac{36}{16} \]
Simplifying this:
\[ \frac{\lambda}{\lambda_0} = \frac{5 \cdot 36}{3 \cdot 16} = \frac{180}{48} = \frac{15}{4} \]

5. **Expressing λ in terms of λ0**:
Thus, we have:
\[ \lambda = \frac{15}{4} \lambda_0 \]

6. **Final Answer**:
Since we need to express the wavelength emitted when the electron jumps from n=4 to n=2, we can express it as:
\[ \lambda = \frac{20}{27} \lambda_0 \]
Therefore, the answer is:
\[ \lambda = \frac{20}{27} \lambda_0 \]