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 will follow these steps:

### Step 1: Understand the Excitation and De-excitation Process
The electron in the lithium ion \( \text{Li}^{++} \) is initially in the ground state (n=1) and is excited to a higher energy level (n=n2) by radiation of wavelength \( \lambda \). When the ion de-excites back to the ground state, it can do so in multiple ways, emitting photons and resulting in spectral lines.

### Step 2: Determine the Number of Spectral Lines
The total number of spectral lines observed during the de-excitation process is given as 6. The formula for the number of spectral lines \( L \) that can be observed when an electron transitions from level \( n_2 \) to level \( n_1 \) is given by:
\[ L = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2} \]
Since the electron is transitioning from \( n_2 \) to \( n_1 = 1 \), we can simplify this to:
\[ L = \frac{(n_2 - 1)(n_2)}{2} \]
Setting \( L = 6 \):
\[ \frac{(n_2 - 1)(n_2)}{2} = 6 \]

### Step 3: Solve for \( n_2 \)
Multiplying both sides by 2:
\[ (n_2 - 1)(n_2) = 12 \]
Expanding this gives:
\[ n_2^2 - n_2 - 12 = 0 \]
This is a quadratic equation. We can solve it using the quadratic formula:
\[ n_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -1, c = -12 \):
\[ n_2 = \frac{1 \pm \sqrt{1 + 48}}{2} = \frac{1 \pm 7}{2} \]
Calculating the two possible values:
\[ n_2 = 4 \quad \text{(valid, since n must be positive)} \]
\[ n_2 = -3 \quad \text{(not valid)} \]

### Step 4: Calculate the Energy Difference
Now that we have \( n_2 = 4 \), we can calculate the energy difference between the two levels using the formula for the energy of an electron in a Bohr orbit:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
For \( \text{Li}^{++} \), \( Z = 3 \):
- Energy at \( n=4 \):
\[ E_4 = -\frac{13.6 \times 3^2}{4^2} = -\frac{13.6 \times 9}{16} = -7.65 \text{ eV} \]
- Energy at \( n=1 \):
\[ E_1 = -\frac{13.6 \times 3^2}{1^2} = -13.6 \text{ eV} \]

### Step 5: Calculate the Energy Difference
The energy difference \( \Delta E \) when the electron transitions from \( n=4 \) to \( n=1 \):
\[ \Delta E = E_1 - E_4 = -13.6 - (-7.65) = -5.95 \text{ eV} \]

### Step 6: Relate Energy to Wavelength
Using the relation between energy and wavelength:
\[ E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{E} \]
Substituting the values:
- \( h = 6.63 \times 10^{-34} \text{ Js} \)
- \( c = 3 \times 10^8 \text{ m/s} \)
- \( E = 5.95 \text{ eV} \) (convert eV to Joules: \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \)):
\[ E = 5.95 \times 1.6 \times 10^{-19} \text{ J} = 9.52 \times 10^{-19} \text{ J} \]
Now, substituting into the wavelength formula:
\[ \lambda = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{9.52 \times 10^{-19}} \approx 10.8 \times 10^{-9} \text{ m} = 10.8 \text{ nm} \]

### Final Answer
The value of \( \lambda \) is approximately \( 10.8 \text{ nm} \).

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