A sample of hydrogen gas in its ground state is irradiated with photons of `10.2 eV` energies. The radiation from the above sample is used to irradiate two other sample of excited ionized `He^(+)` and excited ionized `Li^(2+)`, repectively. Both the ionized samples absorb the incident radiation.
What is the smallest wavelength that will be observed in spectra of `He^(+)` ion ?

Options

A. `24.4 nm`
B. `28.8 nm`
C. `22.2 nm`
D. `30.6 nm`

Explanation

To solve the problem, we need to determine the smallest wavelength that can be observed in the spectra of the ionized helium ion (He⁺) after it absorbs the radiation of 10.2 eV energy.

### Step-by-Step Solution:

1. **Identify the Energy Levels of He⁺:**
The energy levels of a hydrogen-like atom (like He⁺) are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( Z \) is the atomic number. For He⁺, \( Z = 2 \):
\[ E_n = -\frac{13.6 \cdot 2^2}{n^2} = -\frac{54.4}{n^2} \, \text{eV} \]

2. **Calculate the Energy Levels:**
- For \( n = 1 \):
\[ E_1 = -54.4 \, \text{eV} \]
- For \( n = 2 \):
\[ E_2 = -13.6 \, \text{eV} \]
- For \( n = 3 \):
\[ E_3 = -6.04 \, \text{eV} \]
- For \( n = 4 \):
\[ E_4 = -3.4 \, \text{eV} \]

3. **Determine the Transition Caused by Absorption of 10.2 eV:**
The energy of 10.2 eV is absorbed by an electron in the He⁺ ion. We need to find which transition corresponds to this energy. The possible transitions are:
- From \( E_2 \) to \( E_4 \):
\[ \Delta E = E_4 - E_2 = (-3.4) - (-13.6) = 10.2 \, \text{eV} \]
This indicates that the electron transitions from \( n = 2 \) to \( n = 4 \).

4. **Calculate the Energy for the Smallest Wavelength:**
The smallest wavelength corresponds to the transition from the highest energy level (n=4) to the lowest energy level (n=1):
\[ \Delta E = E_4 - E_1 = (-3.4) - (-54.4) = 51 \, \text{eV} \]

5. **Convert Energy to Wavelength:**
We use the relation between energy and wavelength:
\[ E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{E} \]
where \( h = 6.63 \times 10^{-34} \, \text{J s} \) and \( c = 3 \times 10^8 \, \text{m/s} \). First, convert \( E \) from eV to Joules:
\[ E = 51 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 8.16 \times 10^{-18} \, \text{J} \]
Now substituting the values:
\[ \lambda = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{8.16 \times 10^{-18} \, \text{J}} \approx 2.43 \times 10^{-7} \, \text{m} = 24.3 \, \text{nm} \]

### Final Answer:
The smallest wavelength that will be observed in the spectra of He⁺ ion is approximately **24.3 nm**.