If k be the kinetic energy and m the mass of a moving particle then the de-Broglie wavelength of the particle is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \lambda = \displaystyle\frac{h}{\sqrt{2mk}} \)
The de Broglie wavelength of a particle is Planck's constant divided by its momentum, \( \lambda = h/p \). The momentum follows from the kinetic energy:
\( \displaystyle k = \frac{p^{2}}{2m} \;\Rightarrow\; p = \sqrt{2mk} \)
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mk}} \)
Note: The same question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1).