Monochromatic radiation of wave length 640.2 nm `(1nm=10^(-9)m)` from a neon lamp irradiates a photosensitive material made of calcium or tungsten. The stopping voltage is measured to be 0.54V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photocell. Predict the new stopping voltage.

Explanation

Wavelength of the monochromatic radiation, `lambda =640.2nm= 640.2xx10^(-9)m`
Stopping potential of the neon lamp, `V_(0)=0.54V`
Charge on an electron `e=1.6xx10^(-19)C`
Planck's constant, `h=6.6xx10^(-34)Js`
Let `phi_(0)` be the work function and v the frequency of emitted light. We have the photo-energy relation form the photoelectric effect as :
`eV_(0)=hv-^(phi_(0)`
`phi_(0)=(hc)/(lambda)-eV_(0)`
`=(6.6xx10^(-34)xx3xx10^(8))/(640.2xx10^(-9))-1.6xx10^(-10)xx0.54`
`=3.093xx10^(-19)-0.864xx10^(-19)`
`=2.229xx10^(-19)J`
`(2.229xx10^(19))/(1.6xx10^(-19))=1.39eV`
Wavelength of the radiation emittted from an iron source,` lambda' =427.2 nm =427.2xx10^(-9)m`
Let`.^V0` be the new stopping potential, Hence, photo -energy is given as : `.eV_0'(hc)/(lambda')-phi_(0)`
`=(6.6xx10^(-34)xx3xx10^(8))/(427.2xx10^(-9))-2.229xx10^(-19)`
`=4.63xx10^(-19)-2.229xx10^(-19)`
`=2.401xx10^(-19)J`
`=(2.401xx10^(-19))/(1.6xx10^(-19))=1.5eV`
Hence, the new stopping potentical is `1.50eV`.