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The threshold frequency ? 0 for a metal is 7.0 ×1014 s–1. Calculate the kinetic energy of an electron emitted when radiation of frequency ? =1.0 ×1015 s–1 hits the metal.

sudhanshu , 11 Years ago
Grade 12
anser 2 Answers
Gaurav

Last Activity: 10 Years ago

Hello Student
According to Einstein's equation Kinetic energy = 1/2 mev2=h(v - vo)
=(6.626 x104J s) (10.0 x10-14- 7.0 x10" s-1)
= 1.988 x 10-19 J
Saurabh bhatt

Last Activity: 8 Years ago

Applying Einstein`s kinetic energy equation = 1/2mv^2 = h(v-vo).h=6.626 x 10-34 joule-seconds.V=1 x 10^15 s^-1.Vo=7 x 10^14 s^-1.Kinetic energy = (6.626 x 10-34 J s )(10 x 10^14 - 7 x 10^14).=(6.626 x 10-34) x (3 x 10^14).=1.9878 x 10-20...........##........##.........##.........##........##.....##
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