Nuclear Physics and RadioactivityHard
Question
When photons of energy 4.25 eV strike the surface of a metal A, the ejected photo electrons have maximum kinetic energy TA eV and de-Broglie wavelength λA. the maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA - 1.5 eV). If the de-Broglie wavelength of these electrons of these electrons are related as λB = 2λA = , then
Options
A.The work function of A is 2.25 eV
B.The work function of B is 4.20 eV
C.TA = 2.00 eV
D.TB = 2.75 eV
Solution
(1) λA = 
(2) λB = 
(3) TB = TA - 1.5 eV (4) λB = 2λA
(2) λB = (3) TB = TA - 1.5 eV (4) λB = 2λA
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