Work, Power and EnergyHard
Question
In PEE, 4 eV is the energy of the incident photon and the work function is 2 eV. The minimum de-Broglie wavelength of the emitted photoelectrons is :-
Options
A.3√5 (Ao)
B.2 (Ao)
C.5√3 (Ao)
D.4 (Ao)
Solution
= W0 + eV0 ⇒ 4 eV = 2 eV + eV0
⇒ V0 = 2 volt
⇒ λ =
(Ao) = 5√3(Ao)
⇒ V0 = 2 volt
⇒ λ =
(Ao) = 5√3(Ao)Create a free account to view solution
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