If k be the kinetic energy and m is the mass of a moving particle, th…

The correct answer is Option (4) → \( \lambda=\displaystyle\frac{h}{\sqrt{2mk}} \)

The de Broglie wavelength is \( \lambda = h/p \). The momentum follows from the kinetic energy:

    \( \displaystyle k = \frac{p^{2}}{2m} \)

    \( \displaystyle p = \sqrt{2mk} \)

    \( \displaystyle \lambda = \frac{h}{\sqrt{2mk}} \)

Note: The same question was also asked in CUET (UG) 2023 on 26-05-2023 (Slot 1).

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), but it is still in the CUET 2026 syllabus, so do not skip it.

NCERT (Pre-rationalised edition, 2022-23):
Wave Nature of Matter – “… Consider an electron (mass m, charge e) accelerated from rest through a potential V. The kinetic energy K of the electron equals the work done (eV) on it by the electric field:

    \( \displaystyle K = eV \qquad (11.8) \)

Now, \( K = \frac{1}{2}mv^2 = \frac{p^2}{2m} \), so that

    \( \displaystyle p = \sqrt{2mK} = \sqrt{2meV} \qquad (11.9) \)

The de Broglie wavelength \( \lambda \) of the electron is then

    \( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2meV}} \qquad (11.10) \)”