For a monochromatic light incident on a metal surface, the maximum velocity of the emitted photoelectrons is v. Then the stopping potential would be

Options

A. `(2mv^(2))/(e)`
B. `(mv^(2))/(e)`
C. `(mv^(2))/(2e)`
D. `(mv^(2))/(sqrt2e)`

Answer

C

Explanation

If `V_(0)` is the stopping potential, then maximum kinetic energy of emitted photoelectrons `= eV_(0)`
`:. eV_(0) = (1)/(2) mv^(2)`
or, `V_(0) = (mv^(2))/(2e)`