H-atoms is exposed to electromagnetic radition of `1028Å`and gives out induced raditions ( raditions
emitted wher `e^(-)` return to ground state ) .Calculate `lambda` of induced radiations.

Explanation

To calculate the wavelength of the induced radiations emitted by a hydrogen atom when it returns to the ground state after being exposed to electromagnetic radiation of 1028 Å, we can follow these steps:

### Step 1: Convert the Wavelength from Angstroms to Meters
The given wavelength is in angstroms (Å). We need to convert it to meters for our calculations.

\[ \lambda = 1028 \, \text{Å} = 1028 \times 10^{-10} \, \text{m} \]

### Step 2: Use the Rydberg Formula
The Rydberg formula for hydrogen 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, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( n_1 \) is the principal quantum number of the lower energy level (ground state, \( n_1 = 1 \))
- \( n_2 \) is the principal quantum number of the higher energy level

### Step 3: Substitute Known Values into the Rydberg Formula
Substituting \( n_1 = 1 \) into the Rydberg formula:

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

### Step 4: Calculate \( n_2 \)
Using the wavelength of the incident radiation, we can find \( n_2 \):

\[ \frac{1}{1028 \times 10^{-10}} = 1.097 \times 10^7 \left( 1 - \frac{1}{n_2^2} \right) \]

Calculating the left side:

\[ \frac{1}{1028 \times 10^{-10}} \approx 9.73 \times 10^6 \]

Setting the equation:

\[ 9.73 \times 10^6 = 1.097 \times 10^7 \left( 1 - \frac{1}{n_2^2} \right) \]

### Step 5: Solve for \( n_2 \)
Rearranging gives:

\[ 1 - \frac{1}{n_2^2} = \frac{9.73 \times 10^6}{1.097 \times 10^7} \]

Calculating the right side:

\[ 1 - \frac{1}{n_2^2} \approx 0.886 \]

Thus:

\[ \frac{1}{n_2^2} = 1 - 0.886 = 0.114 \]

Taking the reciprocal gives:

\[ n_2^2 \approx \frac{1}{0.114} \approx 8.77 \]

Taking the square root gives:

\[ n_2 \approx 3 \]

### Step 6: Calculate the Wavelength of Induced Radiation
Now, we can calculate the wavelength of the induced radiation when the electron transitions from \( n_2 = 3 \) to \( n_1 = 1 \):

Using the Rydberg formula again:

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

Calculating:

\[ \frac{1}{\lambda_{\text{induced}}} = 1.097 \times 10^7 \left( 1 - \frac{1}{9} \right) = 1.097 \times 10^7 \left( \frac{8}{9} \right) \]

Calculating the right side:

\[ \frac{1}{\lambda_{\text{induced}}} \approx 9.73 \times 10^6 \]

Thus:

\[ \lambda_{\text{induced}} \approx \frac{1}{9.73 \times 10^6} \approx 1.03 \times 10^{-7} \, \text{m} = 1030 \, \text{Å} \]

### Final Answer
The wavelength of the induced radiation is approximately **1030 Å**.

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