Question

The energy of an electron having de-Broglie wavelength ${}^{\prime }{\lambda }^{\prime }$ is:

[h = Planck's constant, m = mass of electron]

• A. $\frac{h}{2m\lambda }$
• B. $\frac{h^2}{2m{\lambda }^2}$
• C. $\frac{h^2}{2m^2{\lambda }^2}$
• D. $\frac{h^2}{2m^2\lambda }$

Answer 1

Answer: (B)

$\frac{h^2}{2m{\lambda }^2}$

$\lambda =\frac{h}{mv}$

$⟹v=\frac{h}{m\lambda }$

$K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\frac{h}{m\lambda }\right)}^2$

$⟹K=\frac{h^2}{2m{\lambda }^2}$

Hence, the answer is option-(B).