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

Wavelength, `lambda = 6000 Å = 6000 xx 10^(-10) m`
stopping potential, `V_(0) = 0.8 V`
(i) Frequency of the light, `upsilon = (c)/(lambda)`
`= (3 xx 10^(8))/(6000 xx 10^(-10)) = 5 xx 10^(-4) xx 10^(18)`
`upsilon = 5 xx 10^(14)` Hz
(ii) Energy of the incident photon,
`E = h upsilon = 6.6 xx 10^(-34) xx 5 xx 10^(14)`
`= 33 xx 10^(-20)J`
`= (33 xx 10^(-20))/(1.6 xx 10^(-19)) = 20.625 xx 10^(-1)`
E = 2.06 eV
(iii) Work function of the cathode material,
`W_(0) = h upsilon - e V_(0)`
`= ((6.6 xx 10^(-34) xx 5 xx 1-^(14))/(1.6 xx 10^(-19))) - ((1.6 xx 10^(-19) xx 0.8)/(1.6 xx 10^(-19))) = 2.06 - 0.8`
`W_(0) = 1.26 eV`
(iv) Threshold frequency, `W_(0) = h upsilon_(0)`
`upsilon_(0) = (W_(0))/(h) = (1.26 xx 1.6 xx 10^(-19))/(6.6 xx 10^(-34)) = 0.3055 xx 10^(15)`
`upsilon_(0) = 3.05 xx 10^(14)` Hz
(v) Net energy of the electron after it leaves the surface
`E = (upsilon - upsilon_(0))`
`= 6.6 xx 10^(-34) (5 xx 10^(14) - 3.06 xx 10^(14))`
`= 6.6 xx 10^(-34) xx 1.94 xx 10^(14)`
`E = 12.804 xx 10^(-20) J`
`= (1.2804 xx 10^(-19))/(1.6 xx 10^(-19))`
E = 0.8 eV