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Ultraviolet light of 6.2ev falls on Caesium surface(work function=1.2ev). The kinetic energy (in electron volts) of the fastest electron emitted is approximately..

Ultraviolet light of 6.2ev falls on Caesium surface(work function=1.2ev). The kinetic energy (in electron volts) of the fastest electron emitted is approximately..

Grade:11

1 Answers

Vikas TU
14149 Points
5 years ago
i think......
Ultraviolet light of 6.2ev falls on Caesium surface(work function=1.2ev). The kinetic energy (in electron volts) of the fastest electron emitted is approximately..
 
hv = W  + K.E
k.e = hv – W
K.E = 6.2ev – 1.2 ev = >5  ev.

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