A photoelectric surface is illuminated successively by monochromatic light of wavelength `lambda` and `(lambda)/(2)`.If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times in the first case,the work function of the surface of the material is: (h=Plank's constant ,c=speed of light)

Options

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

Answer

B

Explanation

Maximum kinetic energy K`(hc)/(lambda)-phi`
In first case ,`K_(1)=(hc)/(lambda)-phi`
In second case ,`K_(2)=(hc)/((lambda)/(2))-phi=(2hc)/(lambda)-phi`
But `K_(2)=3K_(1)` is given .
`therefore (2hc)/(lambda)-phi=(3hc)/(lambda)-3phi`
`3phi-phi=(3hc)/(lambda)-(2hc)/(lambda)`
`2phi=(hc)/(lambda)`
`therefore phi=(hc)/(2lambda)`