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
View Solution FreeMore Work, Power and Energy Questions
Position of a particle of mass 2 kg is given by x = + t2, then what will be the work done in first two seconds :-...An object of mass 5 kg falls from rest through a vertical distance of 20 m and reaches ground with a velocity of 10 m/s....At ordinary temperatures, the molecules of an ideal gas have only translational and rotational kinetic energies. At high...A block of mass 2 kg is free to move along the x-axis. It is at rest and from t = 0 onwards it is subjected to a time-de...The unit of the co-efficient of restitution is -...