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

Options

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

Answer

B

Explanation

According to de-progdie hypothesis
`lamda=(h)/(mv) therefore v= (h)/(m lamda) `
Now Energy of electron
`E= (1)/(2) mv^(2) =1/2 m (h^(2))/(m^(2)lamda^(2))= (h^(2))/(2mlamda^(2))`