Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut - off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.

Explanation

Given, `lambda = 488 nm = 488xx10^(-9)m, V_(0)=0.38 V, e=1.6xx10^(-19)c, h=6.63xx10^(-34)J-s`
`c=3xx10^(8)m//s rArr KE = eV_(0)=(hc)/(lambda) - phi_(0) rArr 1.6xx10^(-19)xx0.38=(6.63xx10^(-34)xx3xx10^(8))/(488xx10^(-9))- phi_(0)`
`6.08xx10^(-20)=40.75xx10^(-20)-phi_(0) rArr phi_(0) = (40.75-6.08)xx10^(-20)=34.67xx0^(-20)J`
`= (34.67xx10^(-20))/(1.6xx10^(-19))eV therefore phi_(0)=2.17 eV`.