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

Options

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

Answer

B

Explanation

The kinetic energy of the electron emitted from the metal surface will be
`(1)/(2) mv^(2)= eV rArr p^(2)=2meV " Hence", sqrt(2meV) =(h)/(lambda) rArr lambda =(h)/(sqrt(2meV))`