InterviewSolution
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InterviewSolution
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Ultraviolet light of wavelength `2271Å` from a 100W mercury souce irradiates a photocell made of molybdenum metal. If the stopping potential is 1.3V, estimate the work function of the metal. How would the photocell respond to a high intensity `(~10^(5)Wm^(-2))` red light of wave length `6328Å` produced by a He-Ne laser? |
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Answer» Here, `lambda=2271Å=2271xx10^(-10)m, V_(0)=1.3V, phi_(0)=?` we know, `eV_(0)=hv-phi_(0) or phi_(0)=hv-eV_(0)=(hc)/lambda-eV_(0) = (6.63xx10^(-34)xx3xx10^(8))/(2271xx10^(-10))-1.6xx10^(-19)xx1.3` `phi_(0)=6.7xx10^(-19)J= (6.7xx10^(-19))/(1.6xx10^(-19))eV=4.2 eV` If `V_(0)` is the frequency, then `hv_(0) =(phi_(0))/h=(6.7xx10^(-19))/(6.63xx10^(-34)) =1.0xx10^(15)Hz` For red light, `lambda_(r)=6328Å=6238xx10^(-10)m`. we have, frequency `v_(r)=c/(lambda_(r))=(3xx10^(8))/(6238xx10^(-10))=4.74xx10^(14)Hz`. Since, `v_(r) lt v_(0)`, therefore, the photocell will not respond to this red light, however strong its intensity may be. |
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