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
frequency of the light

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 = ?
Frequency of incident light `v = ( c )/( lambda)`
Velocity of light `c = 3 xx 10^(8) m//s `
`v = ( 3 xx 10^(8))/( 6 xx 10^(-7)) = 0.5 xx 10^(15) = 5 xx 10^(14) Hz`
`v = 5 xx 10^(14) Hz`