1.

Light of wavelength 5000Å falls on a metal surface of work function 1.9eV. Find (i) the energy of photons in eV (ii) the kinetic energy of photoelectrons and (iii) the stopping potential. Use `h=6.63xx10^(-34)Js`, `c=3xx10^8ms^-1 , e=1.6xx10^-9C.`

Answer» Here, `lambda=5000Å=5xx10^-7m`,
`phi_0=1.9eV`
(i) Energy of photon in eV,
`E=(hc)/lambda=((6.63xx10^(-34))xx(3xx10^8))/((5xx10^-7)xx(1.6xx10^(-19)))eV`
`=2.4825eV`
(ii) Kinetic energy of photoelectron
`=(hc)/lambda-phi_0=2.4825-1.9`
`=0.5825eV`
(iii) If `v_0` is the stopping potential, then
`eV_0=K.E.` of emitted photoelectron =0.5825eV
or `V_0=(0.5825eV)/e=0.5825V`


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