When light of frequency `6 xx 10^(14)` Hz is incident on a photosensitive metal, the kinetic energy of the photoelectron ejected is 2e V. Calculate the kinetic energy of the photoelectron in eV when light to frequency `5 xx 10^(14)` Hz is incident on the same metal.
Given Planck's constant, `h = 6.625 xx 10^(-34) Js`, `1 eV = 1.6 xx 10^(-19)`.

Explanation

`hv = omega + K.E.`
`(6.625 xx 10^(-34) xx 6 xx 10^(14))/(1.6 xx 10^(-19)) = omega + 2eV` ..(i)
`(6.625 xx 10^(-34) xx 5 xx 10^(14))/(1.6 xx 10^(-19)) = omega + K.E.` ...(ii)
(i) - (ii)`implies K.E. = 1.5859 eV`
Detailed Answer:
According to Einsteins photoelectric equation,
hv = W + KE where, W = Work function for incident frequency `v = 6 xx 10^(14) Hz`,
`(6.625 xx 10^(-34) xx 6 xx 10^(14))/(1.6 xx 10^(-19))eV = W + 2 eV` ...(i)
also for frequency `5 xx 10^(14)` Hz,
`(6.625 xx 10^(-34) xx 5 xx 10^(14))/(1.6 xx 10^(-19))eV = W + K.E.` ...(ii)
Subtracting (ii) -(i)
`(6.625 xx 10^(-34) xx 10^(14))/(1.6 xx 10^(-19))[5 - 6] eV = K.E. - 2 eV`
-0.414 eV = K.E. - 2 eV
K.E. = 2 - 0.414
= 1.585 eV