When a `6000Å` light falls on the cathode of a photo cell and produced photoemission. If a stopping potential of 0.8 V is required to stop emission of electron, then determine the
energy of the incident photon

Explanation

Given data `:`
Wavelength of incident light `lambda =6000Å = 6 xx 10^(-7) m `
Stopping potential `V_(0) =0.8 V`
To find `:`
(i) Frequency of light v = ?
(ii) Energy of incident photon E =?
(iii) Work function of the cathode material W= ?
(iv) Threshold frequency `v_(0) = ?`
(v) Net energy of the electron after it leaves the surface E = ?
Energy of the incident photon `E = hv `
Planck's constant `h = 6.626 xx 10^(-34)`
`E = 6.626 xx 10^(-34) xx 5 xx 10^(14) = 33.13 xx 10^(-20) J`
`E = ( 33 xx 10^(-20))/( 1.6 xx 10^(-19)) = 20.76 xx 10^(-1) = 2.07 eV`
` E = 2.07 eV`