JEE Chemistry 2026 Question with Solution

Given: The spectral line is for $\text{Li}^{2+}$, so this is a hydrogen-like species with $Z = 3$. The two levels satisfy $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$.

Find: The wavelength $\lambda$ of the emitted spectral line.

From the two conditions,

$$\begin{aligned} n_1 + n_2 &= 4 \\ n_2 - n_1 &= 2 \end{aligned}$$

Adding the equations gives

$$2n_2 = 6$$

so

$$n_2 = 3$$

and therefore

$$n_1 = 1$$

Now use the Rydberg formula for a hydrogen-like ion:

$$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

Substituting $R = 1.097 \times 10^5 \, \text{cm}^{-1}$, $Z = 3$, $n_1 = 1$, and $n_2 = 3$,

$$\frac{1}{\lambda} = 1.097 \times 10^5 \times 9 \left(1 - \frac{1}{9}\right)$$ $$\frac{1}{\lambda} = 1.097 \times 10^5 \times 8$$

Hence,

$$\lambda = 1.14 \times 10^{-6} \, \text{cm}$$

Therefore, the correct option is D.

Given: The sum of the two principal quantum numbers is $4$ and their difference is $2$.

Find: The wavelength corresponding to the transition in $\text{Li}^{2+}$.

First identify the levels. The only pair satisfying both conditions is

$$(n_1, n_2) = (1, 3)$$

For hydrogen-like species, wavelength depends on $Z^2$, so for $\text{Li}^{2+}$ we use $Z = 3$. The transition formula is

$$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

Thus,

$$\frac{1}{\lambda} = R \times 9 \times \left(1 - \frac{1}{9}\right) = 8R$$

Using $R = 1.097 \times 10^5 \, \text{cm}^{-1}$,

$$\frac{1}{\lambda} = 8.776 \times 10^5 \, \text{cm}^{-1}$$

So,

$$\lambda = \frac{1}{8.776 \times 10^5} = 1.14 \times 10^{-6} \, \text{cm}$$

Therefore, the wavelength is $1.14 \times 10^{-6} \, \text{cm}$, so the correct option is D.