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When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, `T_A` (expressed in eV) and deBroglie wavelength `lambda_A`. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20V is `T_B = T_A -1.50eV`. If the deBroglie wavelength of those photoelectrons is `lambda_B = 2lambda_A` thenA. the work function of A is 2.25 eVB. the work function of B is 3.70 eVC. `T_(A)=2.00 eV`D. `T_(B)=2.75 eV` |
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Answer» Correct Answer - ABC `T_(A) =4.25 xxphi_(A) " " lambda_(B)=2lambda_(A)` `T_(B)=4.2 xxphi_(B) " " h/(sqrt(2mT_(B)))=2xxh/(sqrt(2mT_(A)))` `lambda_(A) =1/(sqrt(2mT_(A))), lambda_(B)=1/(sqrt(2mT_(B)))" " 1/(T_(B))=4/(T_(A))` `T_(A)=4T_(B)` `T_(B)=T_(A)-1.5" " T_(A)=2` `T_(B)=4T_(B)-1.5 " " 2=4.25xxphi_(A)` `3T_(B)=1.5 " " phi_(A)=4.25-2=2.25 eV` `T_(B)=0.5 eV` |
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