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

Answer

To solve the problem, we need to determine the wavelength \( \lambda \) of the radiation that excites an electron in the lithium ion \( \text{Li}^{++} \) from the first Bohr orbit to a higher energy level. We know that after excitation, the ion can de-excite in multiple ways, resulting in six spectral lines.

### Step-by-Step Solution:

1. **Identify the Energy Levels**:
The number of spectral lines observed when an electron transitions between energy levels is given by the formula \( \frac{n(n-1)}{2} \), where \( n \) is the number of energy levels involved. We are told that there are 6 spectral lines, so we can set up the equation:
\[ \frac{n(n-1)}{2} = 6 \]
Solving for \( n \):
\[ n(n-1) = 12 \]
The possible values for \( n \) that satisfy this equation are \( n = 4 \) (since \( 4 \times 3 = 12 \)).

2. **Determine the Excited State**:
The electron in the first orbit (ground state, \( n=1 \)) is excited to the fourth energy level (\( n=4 \)).

3. **Calculate the Energy of the Transition**:
The energy levels of a hydrogen-like atom are given by the formula:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
For \( \text{Li}^{++} \), \( Z = 3 \). Thus, we calculate the energy for \( n=1 \) and \( n=4 \):
\[ E_1 = -\frac{13.6 \times 3^2}{1^2} = -\frac{13.6 \times 9}{1} = -122.4 \text{ eV} \]
\[ E_4 = -\frac{13.6 \times 3^2}{4^2} = -\frac{13.6 \times 9}{16} = -7.65 \text{ eV} \]

4. **Calculate the Energy Difference**:
The energy difference \( \Delta E \) when the electron transitions from \( n=4 \) to \( n=1 \) is:
\[ \Delta E = E_1 - E_4 = -122.4 \text{ eV} - (-7.65 \text{ eV}) = -122.4 + 7.65 = -114.75 \text{ eV} \]

5. **Convert Energy to Wavelength**:
The energy of the photon emitted during the transition can also be expressed in terms of wavelength:
\[ \Delta E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{\Delta E} \]
Substituting the values of \( h \) and \( c \):
\[ h = 6.63 \times 10^{-34} \text{ Js}, \quad c = 3 \times 10^8 \text{ m/s} \]
First, convert \( \Delta E \) from eV to Joules (1 eV = \( 1.6 \times 10^{-19} \) J):
\[ \Delta E = 114.75 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 1.836 \times 10^{-17} \text{ J} \]
Now substituting into the wavelength formula:
\[ \lambda = \frac{(6.63 \times 10^{-34} \text{ Js})(3 \times 10^8 \text{ m/s})}{1.836 \times 10^{-17} \text{ J}} \approx 1.08 \times 10^{-8} \text{ m} = 10.8 \text{ nm} \]

### Final Answer:
The value of \( \lambda \) is approximately \( 10.8 \) nm.