CBSE Class 11-science Answered
When a certain metal was irradiated with a light of frequency 3.2 * 10 to d power 16 Hz, the photo electrons had twice the K.E as emitted when d same metal was irradiated with light of frequency 2*10 to d power 16 Hz. Calculate the threshold frequency of the metal.
Asked by prernamehta | 03 Oct, 2010, 10:30: AM
Expert Answer
we know that,
E(Total Energy)=Eo(Minimum Energy reqd. for the electron)+K.E(kinetic Energy of Photo electron)
hv=hv0 + 1/2 mv2
6.6*10-34*(3.2 * 1016) =Eo+ K.E
2.112*10-17 =Eo+ K.E......................(1)
6.6*10-34*(2*1016) =Eo+ 2*K.E.
1.32*10-17=Eo+ 2 K.E .......................(2)
1.32*10-17=Eo+ 2 K.E .......................(2)
solving equation (1) and (2):
Eo=2.904*10-17 j
hvo=2.904*10-17
vo=4.4*1016 Hz.
Answered by | 03 Oct, 2010, 05:32: PM
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