Arrange the values of wavelengths of the emitted radiations in ascending order, when the electron in the hydrogen atom makes the transition between the following energy levels characterized by principal quantum number (n).
(A) n = 2 to n = 1
(B) n = 4 to n = 3
(C) n = 3 to n = 1
(D) n = 3 to n = 2
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C), (A), (D), (B)
The wavelength of the emitted radiation follows from the energy difference between the levels: a larger energy gap gives a shorter wavelength. For hydrogen, by the Rydberg formula,
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right) \)
\( \displaystyle \text{(A) } n = 2 \text{ to } n = 1:\ \frac{1}{1^{2}} - \frac{1}{2^{2}} = \frac{3}{4} = 0.75 \)
\( \displaystyle \text{(B) } n = 4 \text{ to } n = 3:\ \frac{1}{3^{2}} - \frac{1}{4^{2}} = \frac{7}{144} \approx 0.049 \)
\( \displaystyle \text{(C) } n = 3 \text{ to } n = 1:\ \frac{1}{1^{2}} - \frac{1}{3^{2}} = \frac{8}{9} \approx 0.889 \)
\( \displaystyle \text{(D) } n = 3 \text{ to } n = 2:\ \frac{1}{2^{2}} - \frac{1}{3^{2}} = \frac{5}{36} \approx 0.139 \)
\( \lambda \) is inversely proportional to these values, so the wavelengths in ascending order are (C), (A), (D), (B).