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Find the difference of kinetic energies of photoelectrons emitted from a surface by light of wavelength 2500Å and 5000Å. `h=6.62xx10^(-34)Js`.

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The correct Answer is:
`3.96xx10^(-19)J`
`DeltaE=hc[(1)/(lambda_(1))-(1)/(lambda_(2))]=(hc(lambda_(2)-lambda_(1)))/(lambda_(1)lambda_(2))`
`(6.63xx10^(-34)xx3xx10^(8)xx(5000-2500)xx10^(-10))/(2500xx5000xx10^(-20))=3.96xx10^(-19)J`
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