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The maximum kinetic energy of a photoelectron liberated from the surface of lithium with work function 2.35 eV by electromagnetic radiation whose electric component varies with time as: E = a [ 1 + cos (2pi f_(1)t) ] cos 2pi f_(2) t( where a is a constant ) is (f_(1) = 3.6 xx 10^(15) Hz, and f_(2) = 1.2 xx10^(15)Hz and planck's constant h= 6.6 xx 10 ^(-34) Js) |
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Answer» `2.64` eV `E = a Pi + cos (2pif_(1)t)cos (2pif_(2)t)` `rArr E = [a cos ( 2pif_(2)t) + a cos (2pif_(1) t) cos (2pif_(2)t)]` `(cos A cos B = 1/2 cos (A+B)- cos (A-B))` `rArr E = a cos (2pif_(2) t) + a/2 cos 2pi ( f_(1) + f_(2))t-a/2 cos 2 pi ( f_(1) - f_(2))t` So, the electric component has 2 sub- components with frequencies are, `f_(2) (f_(1) + f_(2)) and (f_(1) - f_(2))` So, for maximum kinetic ENERGY of PHOTOELECTRON,whe take photon of maximum frequency. HENCE, `E_(max)=(hc_(max))/e = (6.6xx10^(-34)xx(3.6xx10xx1.2xx10^(15)))/(1.6xx10^(-19))` `=19.8 eV` Hence, the maximum kinetic energy, `KE_(max) =E_(max) - W_(0) = 19.8 - 2.35 = 17.45 eV` Hence, the correct option is (d). |
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