Question

The energy of an electron of mass m moving with velocity V and de-Broglie wavelength $\lambda$ is __________. ('h' is Planck's constant)

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

Answer 1

The correct option is B $\frac{h^2}{2m{\lambda }^2}$

Since the electron is not moving in an electric field, its potential energy is zero and its total energy is equal to kinetic energy.
$E=\frac{1}{2}m{V}^2=\frac{1}{2}m{\left(\frac{h}{m\lambda }\right)}^2=\frac{h^2}{2m{\lambda }^2}$