Choose the correct option from the following transitions in which the…

The correct answer is Option (1) → n = 2 to n = 1

In the Bohr model of the hydrogen atom, a jump from level \( n_2 \) to a lower level \( n_1 \) emits a photon whose energy equals the energy difference of the two levels. The larger this difference, the shorter the wavelength:

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

Working out the bracket for each transition:

    \( \displaystyle 2 \rightarrow 1:\quad \frac{1}{1} - \frac{1}{4} = \frac{3}{4} = 0.75 \)

    \( \displaystyle 3 \rightarrow 2:\quad \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \approx 0.139 \)

    \( \displaystyle 4 \rightarrow 3:\quad \frac{1}{9} - \frac{1}{16} = \frac{7}{144} \approx 0.049 \)

    \( \displaystyle 5 \rightarrow 4:\quad \frac{1}{16} - \frac{1}{25} = \frac{9}{400} \approx 0.023 \)

The jump from \( n = 2 \) to \( n = 1 \) has the largest \( 1/\lambda \), so it gives the minimum wavelength. The levels crowd together as \( n \) grows, so jumps between higher neighbouring levels release less energy.

Note: The same question was also asked in CUET (UG) 2023 on 10-06-2023 (Slot 1) and on 08-06-2023 (Slot 3).

NCERT (2026-27):
The Line Spectra of the Hydrogen Atom – “According to the third postulate of Bohr’s model, when an atom makes a transition from the higher energy state with quantum number \( n_i \) to the lower energy state with quantum number \( n_f \) (\( n_f \lt n_i \)), the difference of energy is carried away by a photon of frequency \( \nu_{if} \) such that

    \( \displaystyle h\nu_{if} = E_{n_i} - E_{n_f} \qquad (12.11) \)”