Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect.When light from this spectral (cutoff)potential of photoelectrons is 0.38 V.Find the work function of the material from which the emitter is made.

Explanation

Here `lambda=488nm=488xx10^(-9)m`
`V_(0)=0.38 V`
`h=6.63xx10^(-34)Js,c=3xx10^(8)ms^(-1)`
`phi_(0)=?`
`implies` In Einsteins.s equation of photoelectric effect,
`(1)/(2)mv_(max)^(2)=hv-phi_(0)`
`therefore eV_(0)=hv-phi_(0)[because (1)/(2)mv_(max)^(2)=V_(0)]`
`therefore phi_(0)=(hc)/(lambda)-eV_(0)[because v=(c)/(lambda)]`
`therefore phi_(0)=(hc)/(lambda)-V_(0)to(in eV)`
`therefore phi_(0)=(6.63xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19)xx488xx10^(-9))-0.38`
`therefore phi_(0)=(0.025473xx10^(2)-0.38)eV`
`therefore phi_(0)~~(2.54-0.38)eV`
`therefore phi_(0)=2.16eV or phi_(0)=3.46xx10^(-19)J`