`Li^(2+)` is initially is ground state. When radiation of wavelength `lamda_(0)` incident on it, it emits 6 different wavelengths during de excitation find `lamda_(0)`

Options

A. `1230Å`
B. `510Å`
C. `970Å`
D. `1480Å`

Answer

To solve the problem, we need to determine the wavelength \( \lambda_0 \) of the radiation incident on the lithium ion \( \text{Li}^{2+} \) that causes it to emit 6 different wavelengths during de-excitation. Let's break down the solution step by step.

### Step 1: Understand the de-excitation process
The lithium ion \( \text{Li}^{2+} \) is initially in the ground state, which corresponds to the quantum number \( n = 1 \). When it absorbs energy from radiation of wavelength \( \lambda_0 \), it gets excited to a higher energy level.

### Step 2: Determine the number of emitted wavelengths
The problem states that during de-excitation, the ion emits 6 different wavelengths. The number of different wavelengths emitted when an electron transitions from a higher energy level \( n \) to a lower energy level \( n = 1 \) can be determined using the formula:
\[ \text{Number of wavelengths} = \frac{n(n-1)}{2} \]
Setting this equal to 6, we can solve for \( n \):
\[ \frac{n(n-1)}{2} = 6 \]
Multiplying both sides by 2 gives:
\[ n(n-1) = 12 \]
This is a quadratic equation:
\[ n^2 - n - 12 = 0 \]
Factoring this equation, we get:
\[ (n - 4)(n + 3) = 0 \]
Thus, \( n = 4 \) (we discard \( n = -3 \) since \( n \) must be a positive integer).

### Step 3: Determine the final state
The electron transitions from \( n = 4 \) to \( n = 1 \) during de-excitation.

### Step 4: Use the Rydberg formula
To find the wavelength \( \lambda \) of the emitted radiation when the electron transitions from \( n = 4 \) to \( n = 1 \), we use the Rydberg formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \( R \) is the Rydberg constant, approximately \( 1.1 \times 10^7 \, \text{m}^{-1} \).
- \( Z \) is the atomic number of lithium, which is 3.
- \( n_1 = 1 \) (ground state) and \( n_2 = 4 \) (excited state).

### Step 5: Substitute the values into the formula
Substituting the values into the Rydberg formula:
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \times 3^2 \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \]
Calculating \( 3^2 = 9 \):
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \times 9 \left( 1 - \frac{1}{16} \right) \]
Calculating \( 1 - \frac{1}{16} = \frac{15}{16} \):
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \times 9 \times \frac{15}{16} \]

### Step 6: Calculate \( \lambda \)
Now, calculating the right-hand side:
\[ \frac{1}{\lambda} = \frac{1.1 \times 10^7 \times 135}{16} \]
Calculating \( 1.1 \times 135 = 148.5 \):
\[ \frac{1}{\lambda} = \frac{148.5 \times 10^7}{16} = 9.28125 \times 10^6 \, \text{m}^{-1} \]
Taking the reciprocal to find \( \lambda \):
\[ \lambda = \frac{1}{9.28125 \times 10^6} \approx 1.077 \times 10^{-7} \, \text{m} = 1077 \, \text{nm} \]

### Step 7: Convert to Angstroms
Since \( 1 \, \text{nm} = 10 \, \text{Å} \):
\[ \lambda \approx 970 \, \text{Å} \]

### Final Answer
Thus, the wavelength \( \lambda_0 \) of the incident radiation is approximately \( 970 \, \text{Å} \).

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