Maximum kinetic energy of a photoelectron is E when the wavelength of incident light is `lambda`. If energy becomes four times when wavelength is reduced to one third, then work function of the metal is

Options

A. `(3hc)/(lambda)`
B. `(hc)/(3lambda)`
C. `(hc)/(lambda)`
D. `(hc)/(2lambda)`

Answer

B

Explanation

`KE_(max)= (hc)/(lambda)-phi`
`E = (hc)/(lambda)-phi`...(i)
`4E = (hc)/(lambda) - phi rArr 4E = (3hc)/((lambda)/3) - phi"…….."(ii)`
(i) and (ii)
`rArr 1/4 = ((hc)/(lambda)-phi)/((3hc)/(lambda)-phi)`
`rArr (3hc)/(lambda)-phi=(4hc)/(lambda)-4phi`
`phi = (hc)/(3lambda)`