Light of wavelength `5,500 overset @A` falls on a sensitive plate with work function 1.7 eV .Find energy of photon ,

Explanation

`lambda=5000Å = 5000 xx10^(-10)m`
`lambda=5xx10^(-7)m`
Work function = `phi_0,= 1.9eV`
`h=6.63xx10^(-34)JS^(-1)`
`c=3xx10^(8)ms^(-1)`
(i) Energy of incident photo = `E=hv=(hc)/lamda`
`E=(6.63xx10^(-34)xx3xx10^8)/(5xx10^(-7))`
`=(19.89)/5xx10^(-34+8+7)`
`(3.978xx10^(19))/(1.6xx10^(19))eV`
`1.6xx10^(19)J`
`=2.49eV.`
(ii) K.E. of ejected photoelectrons `E=-phi_0`
`= 2.49 – 1.90 = 0.59 eV`
(iii) Let Ve=Stopping Potential We knows = K.E. of ejected photoelectrons
`eV, = 0.59 eV`
`V_s^s =0-59` volt.