Light of wavelength 488 nm is produced by an atom laser which is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping potential of photoelectrons is 0.38eV. Find the work function of the material from which the cathode is made. given , `h=6.63xx10^(-34)Js, 1eV=1.6xx10^(-19)J`.

Options

A. `1.38 eV`
B. `2.55 eV`
C. `2.17 eV`
D. `2.93 eV`

Answer

C

Explanation

`E = (hC)/(lambda) = (124eV)/(488) = 2.55 eV`
`K_(max) = eV_(0) = e xx 0.38 = 0.38 eV`
`K_(max) = (hC)/(lambda) - phi_(0) = E - phi_(0)`
`phi_(0) = E - K_(max) = (2.55 - 0.38) = 2.17 eV`