Which transition of `Li^(2+)` is associated with same energy change as `n= 6 ` to ` n=4 ` transtion in `He^(+)` ?

Options

A. n=3 to n=1
B. n=8 to n=6
C. n=9 to n=6
D. n=2 to n=1

Answer

To solve the problem of determining which transition of \( \text{Li}^{2+} \) is associated with the same energy change as the transition from \( n=6 \) to \( n=4 \) in \( \text{He}^{+} \), we will follow these steps:

### Step 1: Understand the Energy Change Formula
The energy change associated with an electron transition in a hydrogen-like atom can be calculated using the formula:
\[ \Delta E = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and upper energy levels, respectively.

### Step 2: Calculate Energy Change for \( \text{He}^{+} \)
For the transition from \( n=6 \) to \( n=4 \) in \( \text{He}^{+} \) (where \( Z = 2 \)):
\[ \Delta E_{\text{He}} = R(2^2) \left( \frac{1}{4^2} - \frac{1}{6^2} \right) \]
Calculating the fractions:
\[ \frac{1}{4^2} = \frac{1}{16}, \quad \frac{1}{6^2} = \frac{1}{36} \]
Finding a common denominator (which is 144):
\[ \Delta E_{\text{He}} = R(4) \left( \frac{9}{144} - \frac{4}{144} \right) = R(4) \left( \frac{5}{144} \right) \]
Thus,
\[ \Delta E_{\text{He}} = \frac{20R}{144} = \frac{5R}{36} \]

### Step 3: Set Up the Equation for \( \text{Li}^{2+} \)
For \( \text{Li}^{2+} \) (where \( Z = 3 \)), we need to find the transition that gives the same energy change. Let’s denote the transition as \( n_1 \) to \( n_2 \):
\[ \Delta E_{\text{Li}} = R(3^2) \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
This simplifies to:
\[ \Delta E_{\text{Li}} = R(9) \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 4: Equate the Energy Changes
Setting the two energy changes equal to each other:
\[ R(9) \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{5R}{36} \]
Cancelling \( R \) from both sides:
\[ 9 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{5}{36} \]
Dividing both sides by 9:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{5}{324} \]

### Step 5: Solve for Possible Transitions
We need to find integer values of \( n_1 \) and \( n_2 \) that satisfy this equation. Testing the provided options:

1. **Option 1: \( n_1 = 3, n_2 = 1 \)**
\[ \frac{1}{3^2} - \frac{1}{1^2} = \frac{1}{9} - 1 = \frac{1 - 9}{9} = -\frac{8}{9} \quad \text{(Not valid)} \]

2. **Option 2: \( n_1 = 8, n_2 = 6 \)**
\[ \frac{1}{8^2} - \frac{1}{6^2} = \frac{1}{64} - \frac{1}{36} = \frac{36 - 64}{2304} = -\frac{28}{2304} \quad \text{(Not valid)} \]

3. **Option 3: \( n_1 = 9, n_2 = 6 \)**
\[ \frac{1}{9^2} - \frac{1}{6^2} = \frac{1}{81} - \frac{1}{36} = \frac{36 - 81}{2916} = -\frac{45}{2916} \quad \text{(Not valid)} \]

4. **Option 4: \( n_1 = 2, n_2 = 1 \)**
\[ \frac{1}{2^2} - \frac{1}{1^2} = \frac{1}{4} - 1 = \frac{1 - 4}{4} = -\frac{3}{4} \quad \text{(Not valid)} \]

After checking all options, we find that the transition \( n_1 = 6 \) and \( n_2 = 9 \) gives us the correct energy change.

### Final Answer
The transition in \( \text{Li}^{2+} \) that corresponds to the same energy change as the transition from \( n=6 \) to \( n=4 \) in \( \text{He}^{+} \) is:
**Option 3: \( n=9 \) to \( n=6 \)**.