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Consider a metal exposed to light of wavelength 600nm.The maximum energyof the electron doubles when light of wavelength 400 nm is used .Find the work function in eV. |
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Answer» Solution :Showing maximum kinetic energy `(1)/(2) mv_(MAX)^(2)` by symbol K,according to photoelectric equation , `hv=hv_(0)+k` `therefore (HC)/(lambda)=phi_(0)+k` (i)For `lambda_(1),(hc)/(lamba_(1))=phi_(0)+K_(1)` …….(1) (II) For `lambda_(2),(hc)/(lambda_(2))=phi_(0)+2K_(1)` ......(2) `(because K_(2)=2K_(1))` From equation (1) and (2) `(hc)/(lambda_(2))=phi_(0)+2((hc)/(lambda_(1))-phi_(0))` `therefore phi_(0)=hc((2)/(lambda_(1))-(1)/(lambda_(2)))` `therefore phi_(0)=6.625x10^(-34)xx3xx10^(8)` `xx((2)/(600xx10^(-9))-(1)/(400xx10^(-9)))` `therefore =(6.625xx10^(-34)xx3xx10^(8))((2)/(600)-(1)/(400))` `therefore phi_(0)=(6.625xx10^(-34)xx3xx10^(8))/(10^(-9))((200)/(600xx400))` `=(6.625xx10^(-34)xx3xx10^(8))/(10^(-9))xx(1)/(1200)` `=1.656xx10^(-9)J` `therefore phi_(0)=(1.656xx10^(-19))/(1.6xx10^(-19))eV=1.035 eV` |
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