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
work function of the cathode material

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 = ?
Work function of the material `phi_(0) = hv - eV_(0)`
`= 33.13 xx 10^(-20) - 1.626 xx 10^(-19) xx 0.8`
`= 33.13 xx 10^(-20) -1.3 xx 10^(-19)`
`= 3.31 xx xx 10^(-19) -1.3 xx 10^(-19)`
`= 2.01 xx 10^(-19) J`
`phi_(0) = (2.01 xx 10^(-19))/( 1.6 xx 10^(-19)) eV`
`phi_(0) = 1.26 eV`