The electron in a hydrogen atom emits wavelengths $λ_1, λ_2, λ_3$ and $λ_4$ in the following transitions (A), (B), (C) and (D), respectively. Arrange the transitions in the increasing order of emitted wavelengths.

(A) n=3 to n=2
(B) n=4 to n=3
(C) n=4 to n=2
(D) n=2 to n=1

Choose the correct answer from the options given below:

(D), (C), (A), (B)

The correct answer is Option (4) → (D), (C), (A), (B)

Use the hydrogen atom formula:

$\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$

(A) $n=3 \to n=2$: $\frac{1}{\lambda_1} = R \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R \cdot \frac{5}{36}$

(B) $n=4 \to n=3$: $\frac{1}{\lambda_2} = R \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R \cdot \frac{7}{144}$

(C) $n=4 \to n=2$: $\frac{1}{\lambda_3} = R \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R \cdot \frac{3}{16}$

(D) $n=2 \to n=1$: $\frac{1}{\lambda_4} = R \left(1 - \frac{1}{2^2}\right) = R \cdot \frac{3}{4}$

Decimal approximations:

$\frac{1}{\lambda_1} \approx 0.1389 R$

$\frac{1}{\lambda_2} \approx 0.0486 R$

$\frac{1}{\lambda_3} = 0.1875 R$

$\frac{1}{\lambda_4} = 0.75 R$

Increasing order of emitted wavelengths:

$\lambda_4 < \lambda_3 < \lambda_1 < \lambda_2$