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For a metal the maximum wavelength required for photoelectron emission is 340 nm, find the work function. If the radiation of wavelength 250 nm falls on the surface of the given metal. Find the maximum kinetic energy of emitted photo electrons in eV. Given Planck's constant = 6.625 xx 10^(-34) Js, velocity of light in vacuum is 3 xx 10^(8)ms^(-1). |
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Answer» Solution :`W=(hc)/(lambda_(0))` `W=(6.625 xx 10^(-34) xx 3 xx 10^(8))/(340 xx 10^(-9))` `W=5.84 xx 10^(-34)` or `W=3.65 eV` `E=(hc)/(lambda)` `=(6.625 xx 10^(-34) xx 3 xx 10^(8))/(250 xx 10^(-9))` `=7.95 xx 10^(-19)J` or `E=4.96 eV` `(K.E)_("max")=E-W` `K.E_("max")=2.11 xx 10^(-19)J` or 1.31 eV |
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