Question

The kinetic energy of an electron having de-Broglie wavelength $\lambda$ is ( $h=$ Planck's constant, $m=$ mass of electron)

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

Answer 1

Answer: (B)

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

Answer 2

$\lambda=\frac{h}{\sqrt{2 m K . E}}$ $\lambda^{2}=\frac{h^{2}}{2 m(K . E)}$ $(K . E)=\frac{h^{2}}{2 m \lambda^{2}}$