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.
Which is the smallest wavelength that will be observed in spectra of `Li^(2+)` ?

Options

A. `8.6 nm`
B. `10.4 nm`
C. `12.8 nm`
D. `14.6 nm`

Explanation

To solve the problem, we need to determine the smallest wavelength observed in the spectra of the ionized lithium ion (Li²⁺) after it absorbs photons of energy 10.2 eV. We will follow these steps:

### Step 1: Calculate the energy levels of Li²⁺
The energy of an electron in the nth orbit of a hydrogen-like atom is given by the formula:

\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]

where \( Z \) is the atomic number. For lithium (Li), \( Z = 3 \).

#### Calculation:
- For \( n = 1 \):
\[ E_1 = -\frac{3^2 \cdot 13.6}{1^2} = -\frac{9 \cdot 13.6}{1} = -122.4 \, \text{eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{3^2 \cdot 13.6}{2^2} = -\frac{9 \cdot 13.6}{4} = -30.6 \, \text{eV} \]
- For \( n = 3 \):
\[ E_3 = -\frac{3^2 \cdot 13.6}{3^2} = -\frac{9 \cdot 13.6}{9} = -13.6 \, \text{eV} \]
- For \( n = 4 \):
\[ E_4 = -\frac{3^2 \cdot 13.6}{4^2} = -\frac{9 \cdot 13.6}{16} = -7.65 \, \text{eV} \]
- For \( n = 5 \):
\[ E_5 = -\frac{3^2 \cdot 13.6}{5^2} = -\frac{9 \cdot 13.6}{25} = -4.86 \, \text{eV} \]
- For \( n = 6 \):
\[ E_6 = -\frac{3^2 \cdot 13.6}{6^2} = -\frac{9 \cdot 13.6}{36} = -3.4 \, \text{eV} \]

### Step 2: Determine the transition due to absorbed energy
The energy of the absorbed photon is 10.2 eV. We need to find which transition can occur in Li²⁺ after absorbing this energy.

Starting from the ground state (n=1):
- The energy required to move from \( n=1 \) to \( n=2 \) is:
\[ E_2 - E_1 = -30.6 - (-122.4) = 91.8 \, \text{eV} \quad (\text{not possible}) \]
- The energy required to move from \( n=1 \) to \( n=3 \) is:
\[ E_3 - E_1 = -13.6 - (-122.4) = 108.8 \, \text{eV} \quad (\text{not possible}) \]
- The energy required to move from \( n=1 \) to \( n=4 \) is:
\[ E_4 - E_1 = -7.65 - (-122.4) = 114.75 \, \text{eV} \quad (\text{not possible}) \]
- The energy required to move from \( n=1 \) to \( n=5 \) is:
\[ E_5 - E_1 = -4.86 - (-122.4) = 117.54 \, \text{eV} \quad (\text{not possible}) \]
- The energy required to move from \( n=1 \) to \( n=6 \) is:
\[ E_6 - E_1 = -3.4 - (-122.4) = 119 \, \text{eV} \quad (\text{not possible}) \]

The only possible transition is from \( n=3 \) to \( n=6 \):
\[ E_6 - E_3 = -3.4 - (-13.6) = 10.2 \, \text{eV} \quad (\text{possible}) \]

### Step 3: Calculate the wavelength of emitted photon
The smallest wavelength corresponds to the transition from \( n=6 \) to \( n=1 \) (highest energy transition).

#### Energy difference:
\[ E = E_6 - E_1 = -3.4 - (-122.4) = 119 \, \text{eV} \]

#### Wavelength calculation:
Using the formula \( E = \frac{hc}{\lambda} \):
\[ \lambda = \frac{hc}{E} \]
Where:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- Convert \( E \) from eV to Joules: \( E = 119 \times 1.6 \times 10^{-19} \, \text{J} \)

Calculating \( \lambda \):
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{119 \times 1.6 \times 10^{-19}} \]

Calculating the values:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{1.904 \times 10^{-17}} \approx 1.04 \times 10^{-8} \, \text{m} = 10.4 \, \text{nm} \]

### Final Answer:
The smallest wavelength that will be observed in the spectra of \( Li^{2+} \) is approximately **10.4 nm**.