A monochromatic beam of light is absorbed by a collector of ground state hydrogen atom in such a way that six different wavelengths are observed when hydrogen relaxes back to the ground state. The wavelength of the incident beam is

Options

A. 97 nm
B. 91 nm
C. 68 nm
D. 85 nm

Answer

To find the wavelength of the incident beam absorbed by a ground state hydrogen atom, we can follow these steps:

### Step 1: Understand the Problem
The problem states that when a monochromatic beam of light is absorbed by a ground state hydrogen atom, six different wavelengths are observed as the atom relaxes back to the ground state. This indicates that the atom has transitioned to multiple excited states before returning to the ground state.

### Step 2: Identify the Excited States
The ground state of hydrogen corresponds to n=1. The excited states can be n=2, n=3, n=4, etc. We need to find out how many transitions can occur from these excited states back to the ground state.

### Step 3: Calculate the Number of Transitions
- From n=2 to n=1: 1 transition
- From n=3 to n=1: 2 transitions (n=3 to n=1 and n=3 to n=2)
- From n=4 to n=1: 3 transitions (n=4 to n=1, n=4 to n=2, n=4 to n=3)

If we consider n=4, we can calculate the total transitions:
- From n=4 to n=1
- From n=4 to n=2
- From n=4 to n=3
- From n=3 to n=1
- From n=3 to n=2
- From n=2 to n=1

This gives us a total of 6 different wavelengths observed.

### Step 4: Use the Rydberg Formula
The Rydberg formula for the wavelengths of the emitted light is given by:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \))
- \( n_1 \) is the lower energy level (ground state, \( n_1 = 1 \))
- \( n_2 \) is the higher energy level (which can be \( n = 2, 3, 4 \))

### Step 5: Calculate the Wavelength for n=4
For the transition from n=4 to n=1, we can substitute into the Rydberg formula:

\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = R \left( 1 - \frac{1}{16} \right) = R \left( \frac{15}{16} \right) \]

### Step 6: Solve for Wavelength
Now we can calculate the wavelength:

\[ \frac{1}{\lambda} = R \cdot \frac{15}{16} \]

Substituting the value of \( R \):

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot \frac{15}{16} \]

Calculating \( \lambda \):

\[ \lambda = \frac{16}{15} \cdot \frac{1}{1.097 \times 10^7} \]

Calculating this gives:

\[ \lambda \approx 97 \, \text{nm} \]

### Conclusion
The wavelength of the incident beam is approximately **97 nm**.

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