Work, Power and EnergyHard
Question
When photons of energy 4.25 eV strike the surface of a metal "A", the ejected photoelectrons have maximum kinetic energy, TA expressed in eV and de Broglie wavelength λ. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA - 1.50 eV). If the de Broglie wavelength of these photoelectrons is 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
4.25 = TA + WA4.70 = TB + WBTB = TA - 1.5
λ =
& λA = 
λB ⇒ TA = 4TB
λ =
& λA = λB ⇒ TA = 4TBCreate a free account to view solution
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