The stopping potential of the electrons emitted in a photoelectric experiment is V. The de-Broglie wavelength of the electron when it is emitted from the metal surface will be

Options

A. `h^(2) // sqrt(2emV)`
B. `h//sqrt(2emV)`
C. `h^(2)//2emV`
D. `h//2emV`

Answer

B

Explanation

The kineetic energy of the electron emitted from the metal surface will be `(1)/(2) mv^(2) =eV rArr (p^(2))/(2m) =eV rArr p^(2)=2 me V" Hence, "sqrt(2 me V) = (h)/(lamda) rArr lamda =(h)/(sqrt(2 me V))`