An electron of mass m is moving with kinetic energy k and momentum p. The de-Broglie wavelength associated will be:
(A) \( \lambda=\displaystyle\frac{\sqrt{2mk}}{p} \)
(B) \( \lambda=\displaystyle\frac{p}{h} \)
(C) \( \lambda=\displaystyle\frac{h}{p} \)
(D) \( \lambda=\displaystyle\frac{h}{\sqrt{2mk}} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (C) and (D) only
The de Broglie wavelength of a particle is Planck’s constant divided by its momentum. The momentum is linked to the kinetic energy by \( k = \frac{p^{2}}{2m} \), that is \( p = \sqrt{2mk} \).
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mk}} \)
(A) \( \lambda=\sqrt{2mk}/p \) — Incorrect. Since \( p=\sqrt{2mk} \), this equals 1, which is not a wavelength.
(B) \( \lambda=p/h \) — Incorrect. This is the inverse of the correct relation.
(C) \( \lambda=h/p \) — Correct. It is the de Broglie relation.
(D) \( \lambda=h/\sqrt{2mk} \) — Correct. It is the same relation written with \( p=\sqrt{2mk} \).