JEE Chemistry 2023 Question with Solution

Given: electron in Li$^{2+}$ is excited from $n$ to $(n+1)$ using radiation of energy $1.47 \times 10^{-17} \, \text{J}$.

Find: the value of $n$.

For a hydrogen-like species, the energy difference between two orbits is

$$\Delta E = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

For Li$^{2+}$, $Z = 3$. Here the transition is from $n$ to $(n+1)$, so

$$\Delta E = R_H \times 9 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$$

Substituting the given values,

$$1.47 \times 10^{-17} = 2.18 \times 10^{-18} \times 9 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$$

Since $2.18 \times 9 = 19.62$, this gives

$$1.47 \times 10^{-17} = 1.962 \times 10^{-17} \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$$

So,

$$\frac{1.47}{1.96} = \frac{3}{4} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$$

Now testing $n = 1$,

$$\frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}$$

This satisfies the equation.

Therefore, the value of $n$ is $1$.

Given: $\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{3}{4}$.

Find: $n$.

Check the smallest integer value first:

$$n = 1$$

Then

$$\frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}$$

This matches immediately.

Therefore, the correct value is $1$.