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Light of wavelength `5000 Å` fall on a metal surface of work function `1.9` eV. Find a. The energy of photon b. The kinetic energy of photoelectrons

Answer» Wavelength of light `(lambda) = 5000 Ã… = 5000 xx 10^(-10)m =5 xx 10^(-7)m`.
Work function `(v^(0))=1.9eV=1.9 xx 1.6 xx 10^(-19) J`.
Step I. Energy of photons `E=hv=(hc)/(lambda)=((6.626 xx 10^(-34)Js)xx (3 xx 10^(8) ms^(-1)))/((5xx 10^(-7)m))=3.97 xx 10^(-19)J`.
Step II . Kinetic energy of photoelectrons ltbr. `KE(1//2 mv^(2))=hv-hv^(0)=(3.97 xx 10^(-19)J)-(1.9 xx 1.6 xx 10^(-19)J) `
`=(3.97 xx 10^(19)J)-(3.04 xx 10^(-19)J)=9.3 xx 10^(-20)J`.


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