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 than in the first case , the work function of the surface of the material is
(`h` = Plank's constant , `c` = speed of light )

Options

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

Answer

C

Explanation

Let `phi_0` be the work function of the surface of the material Then ,
According to Einstein.s photoelectric equaiton, the maximum kinetic energy of the emitted photoelectrons in the first case is
`K_("max"_1)=(hc)/lamda-phi_0`
and that in the second case is
`K_("max"_2)=(hc)/(lamda/2)-phi_0=(2hc)/lamda-phi_0 `
But , `K_("max"_2)=3K_("max"_1)` (given)
`:. (2hc)/lamda-phi_0 =3 ((hc)/lamda-phi_0)`
`3phi_0 -phi_0 =3 ((hc)/lamda -phi_0)`
`3phi_0 - phi_0 = (3hc)/lamda - (2hc)/lamda`
`2phi_0 =(hc)/lamda " or " phi_0 = (hc)/(2lamda)`