In `Li^(++)` , electron in first Bohr orbit is excited to a level by a radiation of wavelength `lamda` When the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of `lamda` (Given : `h= 6.63 xx 10^(-34) Js, c = 3 xx 10^(8) ms^(-1)`)

Options

A. 9.4 nm
B. 12.3 nm
C. 11.4 nm
D. 10.8 nm

Explanation

To solve the problem, we need to find the wavelength \( \lambda \) of the radiation that excites an electron in the \( Li^{++} \) ion from the first Bohr orbit to a higher energy level, given that a total of six spectral lines are observed when the ion de-excites back to the ground state.

### Step 1: Understand the number of spectral lines
The number of spectral lines observed during the de-excitation process can be calculated using the formula:
\[ N = \frac{N_2(N_2 - 1)}{2} \]
where \( N_2 \) is the number of energy levels the electron can transition to from the excited state down to the ground state.

Given that \( N = 6 \):
\[ \frac{N_2(N_2 - 1)}{2} = 6 \]

### Step 2: Solve for \( N_2 \)
Multiplying both sides by 2 gives:
\[ N_2(N_2 - 1) = 12 \]
This can be rearranged into a quadratic equation:
\[ N_2^2 - N_2 - 12 = 0 \]

### Step 3: Factor the quadratic equation
To factor the equation, we look for two numbers that multiply to -12 and add to -1. The numbers are -4 and 3. Thus, we can factor it as:
\[ (N_2 - 4)(N_2 + 3) = 0 \]

### Step 4: Find the possible values of \( N_2 \)
Setting each factor to zero gives:
\[ N_2 - 4 = 0 \quad \Rightarrow \quad N_2 = 4 \]
\[ N_2 + 3 = 0 \quad \Rightarrow \quad N_2 = -3 \quad (\text{not valid}) \]
So, \( N_2 = 4 \) is the only valid solution.

### Step 5: Determine the energy levels
In the Bohr model, the energy levels for a hydrogen-like atom (like \( Li^{++} \)) are given by:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \( Z \) is the atomic number. For \( Li^{++} \), \( Z = 3 \).

### Step 6: Calculate the energy difference
The electron is excited from the first level \( n_1 = 1 \) to \( n_2 = 4 \). The energy difference when the electron transitions from \( n_2 \) to \( n_1 \) is:
\[ \Delta E = E_1 - E_4 = \left(-\frac{3^2 \cdot 13.6}{1^2}\right) - \left(-\frac{3^2 \cdot 13.6}{4^2}\right) \]
Calculating this gives:
\[ \Delta E = -\frac{9 \cdot 13.6}{1} + \frac{9 \cdot 13.6}{16} = -122.4 + 8.55 = -113.85 \, \text{eV} \]

### Step 7: Convert energy to wavelength
Using the relation \( E = \frac{hc}{\lambda} \), we can find \( \lambda \):
\[ \lambda = \frac{hc}{E} \]
Converting \( E \) from eV to Joules (1 eV = \( 1.6 \times 10^{-19} \) J):
\[ E = 113.85 \times 1.6 \times 10^{-19} \, \text{J} = 1.818 \times 10^{-17} \, \text{J} \]
Now substituting \( h = 6.63 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{ms}^{-1} \):
\[ \lambda = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{1.818 \times 10^{-17}} \approx 1.096 \times 10^{-6} \, \text{m} = 1096 \, \text{nm} \]

### Final Answer
The value of \( \lambda \) is approximately \( 1096 \, \text{nm} \).