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Question
The stopping potential in an experiment on photoelectric effect is 1.5V. What is the maximum kinetic energy of the photoelectrons emitted? Calculate in Joules.
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Solution 1
We know that KEmax = eV0
V0 = 1.5V
= 1.6 × 10-19 × 1.5
= 2.4 × 10-19 J
KE = 2.4 × 10-19 J
Solution 2
Using Einstein's photoelectric equation,
`("K.E".)_"max" = "hv" - phi = "h"("v" - "v"_@)`
⇒ `("K.E".)_"max" = "h"("v" - "v"_@) = "eV"_@`
Where Vo is the stopping potential.
According to the given data, Vo = 1.5 V
Hence,
`(K.E.)_"max" = "eV"_@ = "e" xx 1.5 "V" = 1.6 xx 10^-19 xx 1.5 "Joules")`
(K.E.)max = 2.4 × 10-19 Joules
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