An electron of mass m is moving with kinetic energy k and momentum p. The de-Broglie wavelength associated will be:
(A) $\lambda=\frac{\sqrt{2 mk}}{p}$
(B) $\lambda=\frac{p}{h}$
(C) $\lambda=\frac{h}{p}$
(D) $\lambda=\frac{h}{\sqrt{2 mk}}$
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 (λ) of a particle is -
$λ=\frac{h}{P}$ [P = Momentum]
Also,
$P=\sqrt{2m_eK}$
[m = mass of particle]
[K = Kinetic energy of particle]
$∴λ=\frac{h}{P}=\frac{h}{\sqrt{2mK}}$