The electron in a hydrogen atom emits wavelengths \( \lambda_1, \lambda_2, \lambda_3 \) and \( \lambda_4 \) 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:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (D), (C), (A), (B)
The emitted wavelength is given by the Rydberg formula; a larger energy jump gives a shorter wavelength.
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right) \)
(A) \( 3 \rightarrow 2 \): \( \displaystyle \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \approx 0.139 \)
(B) \( 4 \rightarrow 3 \): \( \displaystyle \frac{1}{9} - \frac{1}{16} = \frac{7}{144} \approx 0.049 \)
(C) \( 4 \rightarrow 2 \): \( \displaystyle \frac{1}{4} - \frac{1}{16} = \frac{3}{16} \approx 0.188 \)
(D) \( 2 \rightarrow 1 \): \( \displaystyle 1 - \frac{1}{4} = \frac{3}{4} = 0.750 \)
The wavelength is the reciprocal, so the largest value of \( 1/\lambda \) gives the shortest wavelength:
\( \lambda_{D} \lt \lambda_{C} \lt \lambda_{A} \lt \lambda_{B} \)
The increasing order is (D), (C), (A), (B).