. The threshold frequency v0 for a metal
is 7.0 × 1014 s-1. Calculate the kinetic energy of an electron emitted when radiation of frequency v = 1.0 × 1015 s-1hits the metal.
Best Answer
Explanation:
According to Einstein's equation,
Kinetic energy = 1/2mev2= h( v-v0)
h = Planck’s constant = 6.626×10-34J s
v = Frequency
v0 = Threshold frequency
= (6.626×10-34J s)(1.0×1015 s-1 − 7.0×1014 s-1)
= (6.626×10-34 Js)(10.0×1014 − 7.0×1014)
= (6.626×10-34 Js)(3.0×1014s-1)
= 1.988×10-19J
Final Answer:
Hence the kinetic energy of an electron emitted is 1.988×10−19 J.
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