Maximum velocity of photoelectron emitted is `4.8 ms^(-1)`. If e/m ratio of electron is `1.76xx10^(11)Ckg^(-1)`, then stopping potential is given by

Options

A. `(v^(2))/(2(m/e))`
B. `(v^(2))/(2(e/m))`
C. `(v^(2))/((e/m))`
D. `(v^(2))/((m/e))`

Answer

B

Explanation

Given maximum velcity of photoelectron = v
charge of phoroelectron = e
and mass of photoelectron = m
Let the stopping potential of the photoelectron = v,
Then, the maximum kinetic energy
` KE_ ( max ) = eVrArr ( 1 ) / ( 2 ) mv ^ 2 = eV ( because KE= ( 1 ) / ( 2 ) mv ^ 2 ) `
`rArr V = ( v^ 2 m ) /( 2 e ) or V = ( v ^ 2 ) / ( 2 ((e ) / ( m ) ) ) `