If maximum velocity with which an electron can be emitted from a photo cell is `4 xx 10^(8) cm//sec` , the stopping potential is (mass of electron ` = 9 xx -31 kg`)

Options

A. `30 "volt"`
B. `45 "volt"`
C. `59 "volt"`
D. Information is insufficient

Answer

B

Explanation

`v_(max) = 4 xx 108 cm//sec = 4 xx 106 m//sec`.
`:. K_(max) = (1)/(2) mv_(max)^(2) = (1)/(2) xx 9 xx 10^(-31) xx (4 xx 10^(6))^(2)`
`= 7.2 xx 10^(-18) J = 45 eV`.
Hence , stopping potential
`|V_(0)| = (K_(max))/( e) = 45 v o l t`.