Monochromatic radiation of specific wavelength is incident on H-atoms in ground state. H-atoms absorb energy and emit subsequently radiations of six different wavelength. Find wavelength of incident radiations:

Options

A. (a) 9.75nm
B. (b) 50nm
C. (c) 85.8nm
D. (d) 97.25nm

Answer

To find the wavelength of the incident radiation on hydrogen atoms in the ground state that subsequently emit six different wavelengths, we can follow these steps:

### Step 1: Understand the emission of wavelengths
When hydrogen atoms absorb energy, they can transition to higher energy levels. The fact that they emit six different wavelengths indicates that there are six transitions occurring between energy levels.

### Step 2: Use the formula for the number of spectral lines
The number of spectral lines (or emitted wavelengths) can be calculated using the formula:
\[ \text{Number of lines} = \frac{n(n-1)}{2} \]
where \( n \) is the number of energy levels involved in the transitions.

### Step 3: Set up the equation
Since we know that the number of emitted wavelengths is 6, we can set up the equation:
\[ \frac{n(n-1)}{2} = 6 \]

### Step 4: Solve for \( n \)
Multiplying both sides by 2 gives:
\[ n(n-1) = 12 \]
Now, we can test integer values for \( n \):
- For \( n = 4 \):
\[ 4(4-1) = 4 \times 3 = 12 \quad \text{(This works)} \]
Thus, \( n = 4 \).

### Step 5: Determine the energy levels
The hydrogen atom transitions from the \( n = 4 \) state to lower energy states. The ground state of hydrogen corresponds to \( n = 1 \).

### Step 6: Calculate the wavelength using the Rydberg formula
The Rydberg formula for the wavelength of emitted radiation is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant (\( R \approx 1.1 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 7: Substitute the values
For the transition from \( n_2 = 4 \) to \( n_1 = 1 \):
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \]
\[ = R \left( 1 - \frac{1}{16} \right) = R \left( \frac{16 - 1}{16} \right) = R \left( \frac{15}{16} \right) \]

### Step 8: Calculate \( \lambda \)
Substituting the value of \( R \):
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \cdot \frac{15}{16} \]
Calculating this gives:
\[ \frac{1}{\lambda} = \frac{1.1 \times 15 \times 10^7}{16} \approx 1.03125 \times 10^7 \, \text{m}^{-1} \]
Thus,
\[ \lambda = \frac{1}{1.03125 \times 10^7} \approx 9.72 \times 10^{-8} \, \text{m} = 97.25 \, \text{nm} \]

### Conclusion
The wavelength of the incident radiation is approximately **97.25 nm**.