JEE Chemistry 2024 Question with Solution

Given: The electron in $\text{He}^+$ goes from the fifth excited state to the first excited state.

Find: The number of spectral lines obtained.

For hydrogen-like species, the number of spectral lines possible between two levels $n_2$ and $n_1$ is

$$\text{Number of spectral lines} = \frac{(n_2-n_1)(n_2-n_1+1)}{2}$$

The first excited state corresponds to $n_1 = 2$ and the fifth excited state corresponds to $n_2 = 6$.

Substituting,

$$\text{Number of spectral lines} = \frac{(6-2)(6-2+1)}{2}$$ $$= \frac{4 \times 5}{2}$$ $$= 10$$

Therefore, the number of spectral lines obtained is $10$. Hence, the correct option is C.

Given: Transition is from the fifth excited state to the first excited state in $\text{He}^+$.

Find: Total possible spectral lines.

The fifth excited state means the electron is initially at $n=6$, because ground state is $n=1$. The first excited state means the lowest allowed final level here is $n=2$.

So the electron can take place among the levels

$$6, 5, 4, 3, 2$$

This is a set of $5$ levels. If $m$ levels are involved, the number of possible emission lines is

$$\frac{m(m-1)}{2}$$

Here,

$$m = 5$$

So,

$$\text{Number of lines} = \frac{5 \times 4}{2} = 10$$

Therefore, the number of spectral lines obtained is $10$.