The energy of an electron having de-Brogilie wavelength `lamda` is
(hwere, h=Plank's constant, m = mass of electron)

Options

A. `(h)/(2mlamda)`
B. `(h^(2))/(2mlamda^(2))`
C. `(h^(2))/(2m^(2)lamda^(2))`
D. `(h^(2))/(2m^(2)lamda)`

Answer

B

Explanation

We know that,
`lamda=(h)/(sqrt(2m(KE))`
On taking square on both sides, we get
`lamda^(2)=(h^(2))/(2m(KE))`
`KE=(h^(2))/(2mlamda^(2))`