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 determine 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 can follow these steps:

### Step 1: Calculate the Energy Change for the Transition in \( \text{He}^{+} \)

The energy change (\( \Delta E \)) for a transition in a hydrogen-like atom is given by the formula:

\[ \Delta E = -2.18 \times 10^{-18} Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For \( \text{He}^{+} \):
- Atomic number \( Z = 2 \)
- Transition from \( n_2 = 6 \) to \( n_1 = 4 \)

Substituting these values into the formula:

\[ \Delta E = -2.18 \times 10^{-18} \times 2^2 \left( \frac{1}{4^2} - \frac{1}{6^2} \right) \]

Calculating \( Z^2 \):

\[ Z^2 = 2^2 = 4 \]

Calculating \( \frac{1}{4^2} - \frac{1}{6^2} \):

\[ \frac{1}{16} - \frac{1}{36} = \frac{9 - 4}{144} = \frac{5}{144} \]

Now substituting back into the energy change formula:

\[ \Delta E = -2.18 \times 10^{-18} \times 4 \times \frac{5}{144} \]

Calculating this gives:

\[ \Delta E = -2.18 \times 10^{-18} \times \frac{20}{144} = -2.18 \times 10^{-18} \times \frac{5}{36} \approx -3.03 \times 10^{-19} \text{ J} \]

### Step 2: Set Up the Energy Change for \( \text{Li}^{2+} \)

Now we need to find the transition in \( \text{Li}^{2+} \) that gives the same energy change. The atomic number for lithium is \( Z = 3 \).

Using the same formula for \( \text{Li}^{2+} \):

\[ \Delta E = -2.18 \times 10^{-18} \times 3^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Calculating \( Z^2 \):

\[ Z^2 = 3^2 = 9 \]

Setting the energy change equal to that of \( \text{He}^{+} \):

\[ -2.18 \times 10^{-18} \times 9 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = -3.03 \times 10^{-19} \]

### Step 3: Solve for \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \)

Dividing both sides by \( -2.18 \times 10^{-18} \):

\[ 9 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3.03 \times 10^{-19}}{2.18 \times 10^{-18}} \approx 0.1385 \]

Now we can simplify:

\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{0.1385}{9} \approx 0.01539 \]

### Step 4: Find Suitable \( n_1 \) and \( n_2 \)

Now we need to find integers \( n_1 \) and \( n_2 \) such that:

\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \approx 0.01539 \]

Let's try some values:

- For \( n_1 = 2 \) and \( n_2 = 3 \):

\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \approx 0.1389 \]

This is close, but we need to check other combinations.

- For \( n_1 = 6 \) and \( n_2 = 9 \):

\[ \frac{1}{6^2} - \frac{1}{9^2} = \frac{1}{36} - \frac{1}{81} = \frac{81 - 36}{2916} = \frac{45}{2916} \approx 0.01539 \]

This matches our requirement.

### Conclusion

The transition from \( n=6 \) to \( n=9 \) in \( \text{Li}^{2+} \) is associated with the same energy change as the transition from \( n=6 \) to \( n=4 \) in \( \text{He}^{+} \).