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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..

Yash , 9 Years ago
Grade 11
anser 1 Answers
Vikas TU
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.
Last Activity: 9 Years ago
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