When a certain metal was irradiated with light of frequency 1.6×10^16 hz the photoelectrons emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light of frequency 1.0×10^16 hz . calculate threshold frequency for the metal?
Answers
Answered by
26
According to formula
V=given frequency
V°= threshold frequency
KE = hv-hv°
KE/2kE =1.0×10^6 - v°/1.6×10^6-v°
So after solving the final answer is 4×10^15 hz
V=given frequency
V°= threshold frequency
KE = hv-hv°
KE/2kE =1.0×10^6 - v°/1.6×10^6-v°
So after solving the final answer is 4×10^15 hz
Answered by
2
Answer:
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
Explanation:
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