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`.

Explanation

Wavelength of light produced by the argon laser, `lambda`=488 nm
`=488xx10^(-9) m`
Stopping potential of the photoelectrons, `V_0=0.38 V`
`1eV=1.6xx10^(-19)` J
`therefore V_0=0.38/(1.6xx10^(-19))eV`
Planck's constant, h=`6.6xx10^(-34) Js`
Charge on an electron, `e=1.6xx10^(-19) C`
Speed of light, c=3x10 m/s
From Einstein's photoelectric effect, we have the relation involving the work function `phi_0` of the material of the emitter as :
`eV_0=(hc)/lambda-phi_0`
`phi_0=(hc)/lambda-eV_0`
`=(6.6xx10^(-34)xx3xx10^8)/(1.6xx10^(-19)xx488xx10^(-9))-(1.6xx10^(-19)xx0.38)/(1.6xx10^(-19))`
=2.54-0.38 =2.16 eV
Therefore, the material with which the emitter is made has the work function of 2.16 eV.