A sample of hydrogen gas in its ground state is irradiated with photons of `10.02 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.
How many spectral lines are obtained in the spectra of `Li^(2+)`?

Options

A. 10
B. 15
C. 20
D. 17

Explanation

To solve the problem, we need to determine how many spectral lines are obtained in the spectra of the ionized lithium atom (Li²⁺) when it absorbs radiation of energy 10.02 eV. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the energy levels of Li²⁺
The energy levels of a hydrogen-like atom (such as Li²⁺) can be calculated using the formula:

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

where \( Z \) is the atomic number. For lithium, \( Z = 3 \). Therefore, the energy levels are:

- For \( n = 1 \):
\[ E_1 = -\frac{13.6 \times 3^2}{1^2} = -\frac{13.6 \times 9}{1} = -122.4 \, \text{eV} \]

- For \( n = 2 \):
\[ E_2 = -\frac{13.6 \times 3^2}{2^2} = -\frac{13.6 \times 9}{4} = -30.6 \, \text{eV} \]

- For \( n = 3 \):
\[ E_3 = -\frac{13.6 \times 3^2}{3^2} = -\frac{13.6 \times 9}{9} = -13.6 \, \text{eV} \]

- For \( n = 4 \):
\[ E_4 = -\frac{13.6 \times 3^2}{4^2} = -\frac{13.6 \times 9}{16} = -7.65 \, \text{eV} \]

- For \( n = 5 \):
\[ E_5 = -\frac{13.6 \times 3^2}{5^2} = -\frac{13.6 \times 9}{25} = -4.89 \, \text{eV} \]

- For \( n = 6 \):
\[ E_6 = -\frac{13.6 \times 3^2}{6^2} = -\frac{13.6 \times 9}{36} = -3.4 \, \text{eV} \]

### Step 2: Determine the initial state of Li²⁺
The problem states that Li²⁺ absorbs 10.02 eV of energy. The ground state energy of Li²⁺ is \( E_1 = -122.4 \, \text{eV} \). After absorbing 10.02 eV, the energy of the electron will be:

\[ E = -122.4 + 10.02 = -112.38 \, \text{eV} \]

This energy is higher than the ground state but lower than the first excited state. Therefore, the electron can be excited to \( n = 6 \).

### Step 3: Determine possible transitions
The electron can transition from \( n = 6 \) to any lower energy level (n=1, n=2, n=3, n=4, n=5). The possible transitions are:

- From \( n = 6 \) to \( n = 1 \)
- From \( n = 6 \) to \( n = 2 \)
- From \( n = 6 \) to \( n = 3 \)
- From \( n = 6 \) to \( n = 4 \)
- From \( n = 6 \) to \( n = 5 \)

- From \( n = 5 \) to \( n = 1 \)
- From \( n = 5 \) to \( n = 2 \)
- From \( n = 5 \) to \( n = 3 \)
- From \( n = 5 \) to \( n = 4 \)

- 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 \( n = 1 \)

### Step 4: Count the transitions
To find the total number of unique transitions, we can use the combination formula \( nC2 \) where \( n \) is the number of energy levels the electron can occupy (from n=1 to n=6, which is 6 levels):

\[ \text{Number of transitions} = \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

### Conclusion
The total number of spectral lines obtained in the spectra of Li²⁺ is **15**.

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