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In a photoelectric experiment, when em wave given by E=E0sinwt is incident, electron is just ejected.When E=E0sin2wt is incident Kmax=K1 and when E=E0cos6wt is incident kmax = k2. find k2/k1

In a photoelectric experiment, when em wave given by E=E0sinwt is incident, electron is just ejected.When E=E0sin2wt is incident Kmax=K1 and when E=E0cos6wt is incident kmax = k2. find k2/k1

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1 Answers

vikas askiitian expert
509 Points
11 years ago

E = Eosinwt

frequency(v) = w/2pi so

E = hv = h(w/2pi) = energy of light photon

for this elergy electron is just ejected so this value of Energy is workfunction metal ..

@ = workfunction = h(2pi/w)        .............1

 

for eq E = Eosin2wt , v = 2w/2pi = w/pi

k1 = hv - @ = h(w/pi) - h(w/2pi) = hw/2pi       .............2

for eq E = Eosin6wt , v = 6w/2pi = 3w/pi

K2 = hv - @ = 3wh/pi - wh/2pi = 5wpi/2          ...........3

k2/k1 = 5

 

approve my ans if u like it

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