The wavelength or radiations emitted is `lambda_(0)` when an electron in hydrogen atom jumps from the third orbit to second. If in the H-atom itself, the electron jumps from fourth orbit to second orbit, the wavelength of emitted radiation will be

Options

A. `(20)/(27)lambda_(0)`
B. `(16)/(25)lambda_(0)`
C. `(27)/(20)lambda_(0)`
D. `(29)/(16)lambda_(0)`

Answer

To solve the problem, we will use the Rydberg formula for the hydrogen atom, which relates the wavelengths of emitted radiation to the transitions of electrons between energy levels. 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 radiation,
- \( R \) is the Rydberg constant,
- \( n_1 \) is the principal quantum number of the lower energy level,
- \( n_2 \) is the principal quantum number of the higher energy level.

### Step 1: Calculate the wavelength \( \lambda_0 \) for the transition from the third orbit to the second orbit.

For the transition from \( n_2 = 3 \) to \( n_1 = 2 \):

\[ \frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating the terms:

\[ \frac{1}{\lambda_0} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

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:

\[ \lambda_0 = \frac{36}{5R} \]

### Step 2: Calculate the wavelength \( \lambda \) for the transition from the fourth orbit to the second orbit.

For the transition from \( n_2 = 4 \) to \( n_1 = 2 \):

\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Calculating the terms:

\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) \]

Finding a common denominator (16):

\[ \frac{1}{\lambda} = R \left( \frac{4}{16} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \]

### Step 3: Relate \( \lambda \) to \( \lambda_0 \).

Now substituting \( R \) from the first equation into the second:

From \( \lambda_0 = \frac{36}{5R} \), we can express \( R \) as:

\[ R = \frac{36}{5\lambda_0} \]

Substituting this into the equation for \( \lambda \):

\[ \frac{1}{\lambda} = \frac{36}{5\lambda_0} \cdot \frac{3}{16} \]

This simplifies to:

\[ \frac{1}{\lambda} = \frac{108}{80\lambda_0} = \frac{27}{20\lambda_0} \]

Thus, we find:

\[ \lambda = \frac{20\lambda_0}{27} \]

### Final Answer:

The wavelength of the emitted radiation when the electron jumps from the fourth orbit to the second orbit is:

\[ \lambda = \frac{20\lambda_0}{27} \]