H-atom is exposed to electromagnetic radiation of `lambda=1025.6` Å and excited atom gives out induced radiation. What is the minimum wavelength of the induced radiation?
(a)`102.6 nm`
(b)`12.09` nm
(c)`121.6` nm
(d)`810.8` nm

Options

A. `102.6 nm`
B. `12.09` nm
C. `121.6` nm
D. `810.8` nm

Answer

To solve the problem of determining the minimum wavelength of the induced radiation emitted by a hydrogen atom when exposed to electromagnetic radiation of wavelength \( \lambda = 1025.6 \, \text{Å} \), we will follow these steps:

### Step 1: Convert the wavelength from angstroms to nanometers
The given wavelength is in angstroms (Å), and we need to convert it to nanometers (nm) for our calculations.
1 Å = \( 10^{-1} \) nm, so:
\[ \lambda = 1025.6 \, \text{Å} = 1025.6 \times 10^{-1} \, \text{nm} = 102.56 \, \text{nm} \]

### Step 2: Calculate the energy of the incident radiation
We can use the formula for energy:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 4.135667696 \times 10^{-15} \, \text{eV s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)

Using the value of \( hc \) in electron volts and nanometers:
\[ hc = 1240 \, \text{eV nm} \]
Now substituting the converted wavelength:
\[ E = \frac{1240 \, \text{eV nm}}{102.56 \, \text{nm}} \approx 12.09 \, \text{eV} \]

### Step 3: Determine the possible transitions in the hydrogen atom
The minimum wavelength of the induced radiation corresponds to the maximum energy transition. The possible transitions in hydrogen are:
1. From \( n=3 \) to \( n=1 \)
2. From \( n=2 \) to \( n=1 \)
3. From \( n=4 \) to \( n=2 \)

The transition with the highest energy difference will yield the shortest wavelength.

### Step 4: Calculate the energy difference for the transition from \( n=3 \) to \( n=2 \)
Using the Rydberg formula for hydrogen:
\[ \Delta E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the transition from \( n=3 \) to \( n=2 \):
\[ \Delta E = 13.6 \, \text{eV} \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \, \text{eV} \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the fractions:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus:
\[ \Delta E = 13.6 \, \text{eV} \times \frac{5}{36} \approx 1.89 \, \text{eV} \]

### Step 5: Calculate the wavelength of the emitted radiation
Using the energy calculated:
\[ \lambda = \frac{hc}{\Delta E} = \frac{1240 \, \text{eV nm}}{1.89 \, \text{eV}} \approx 655.03 \, \text{nm} \]

### Step 6: Compare with the options
The minimum wavelength of the induced radiation is approximately \( 655.03 \, \text{nm} \). However, we need to check the transitions from \( n=2 \) to \( n=1 \) and \( n=3 \) to \( n=1 \) for shorter wavelengths.

Calculating for \( n=2 \) to \( n=1 \):
\[ \Delta E = 13.6 \, \text{eV} \left( 1 - \frac{1}{4} \right) = 13.6 \, \text{eV} \times \frac{3}{4} = 10.2 \, \text{eV} \]
\[ \lambda = \frac{1240 \, \text{eV nm}}{10.2 \, \text{eV}} \approx 121.57 \, \text{nm} \]

### Conclusion
The minimum wavelength of the induced radiation is approximately \( 121.6 \, \text{nm} \), which corresponds to option (c).