Radiation coming from transition `n = 2 to n = 1` of hydrogen atoms falls on helium in `n = 1 and n= 2` state what are the possible transition of helium ions as they absorb energy from the radiation?

Explanation

To solve the question about the possible transitions of helium ions when they absorb radiation from the transition of hydrogen atoms from \( n = 2 \) to \( n = 1 \), we can follow these steps:

### Step 1: Calculate the Energy of the Hydrogen Transition
The energy released during the transition from \( n = 2 \) to \( n = 1 \) in a hydrogen atom can be calculated using the formula:

\[ E = 13.6 \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For hydrogen, \( Z = 1 \), \( n_1 = 1 \), and \( n_2 = 2 \):

\[ E = 13.6 \times 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \left( \frac{3}{4} \right) = 10.2 \, \text{eV} \]

### Step 2: Possible Transitions in Helium
Now, we need to find the possible transitions in helium ions (He\(^+\)) when they absorb this energy. The atomic number \( Z \) for helium is \( 2 \).

### Step 3: Calculate Energy for Helium Transitions
Using the same formula for helium transitions, we will calculate the energy for different transitions:

1. **Transition from \( n = 1 \) to \( n = 2 \)**:
\[ E = 13.6 \times 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \times 4 \left( 1 - \frac{1}{4} \right) = 13.6 \times 4 \times \frac{3}{4} = 13.6 \times 3 = 40.8 \, \text{eV} \]

2. **Transition from \( n = 1 \) to \( n = 3 \)**:
\[ E = 13.6 \times 2^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 13.6 \times 4 \left( 1 - \frac{1}{9} \right) = 13.6 \times 4 \times \frac{8}{9} = \frac{108.8}{9} \approx 12.09 \, \text{eV} \]

3. **Transition from \( n = 1 \) to \( n = 4 \)**:
\[ E = 13.6 \times 2^2 \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = 13.6 \times 4 \left( 1 - \frac{1}{16} \right) = 13.6 \times 4 \times \frac{15}{16} = \frac{816}{16} = 51.0 \, \text{eV} \]

4. **Transition from \( n = 2 \) to \( n = 3 \)**:
\[ E = 13.6 \times 2^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \times 4 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \times 4 \left( \frac{5}{36} \right) = \frac{272}{36} \approx 7.56 \, \text{eV} \]

### Step 4: Compare Energies
Now we compare the energies calculated for helium transitions with the energy from the hydrogen transition (10.2 eV):

- The transition from \( n = 1 \) to \( n = 2 \) gives 40.8 eV (too high).
- The transition from \( n = 1 \) to \( n = 3 \) gives approximately 12.09 eV (too high).
- The transition from \( n = 1 \) to \( n = 4 \) gives 51.0 eV (too high).
- The transition from \( n = 2 \) to \( n = 3 \) gives approximately 7.56 eV (too low).

None of the calculated transitions match the energy of 10.2 eV exactly, but the closest transition that can occur is from \( n = 2 \) to \( n = 3 \) as it is the only transition that can absorb energy close to the value of 10.2 eV.

### Conclusion
The possible transition of helium ions as they absorb energy from the radiation is from \( n = 2 \) to \( n = 3 \).

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