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 | 3rd 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)

solving equation (1) and (2):
Eo=2.904*10-17 j

 
hvo=2.904*10-17
 
vo=4.4*1016 Hz.

Answered by  | 3rd Oct, 2010, 05:32: PM

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