The speed of photoelectron is 10^6 ms. What should be the frequency of the
incident radiation on potassium metal whose work function is 2.3 eV?
Answers
Answered by
2
Answer:
E=W
0
+E
k
where, W
0
is the function and E
k
is the KE of th liberated photelectron
W
0
=2.3 eV=3.68 ×10
19
J
E
k
=
2
1
mv
2
=
2
1
×9×10
−31
×(10
4
)
2
J=4.5×10
−23
J
So ,E=(3.68×10
−19
+4.5×10
−23
)J
=3.68045×10
−19
J
Frequency, v=
h
E
=
(6.63×10
−34
)
3.68045×10
19
=0.56×10
15
Hz
Answered By
toppr
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