Saved Bookmarks
| 1. |
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). |
|
Answer» Solution :`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 |
|