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Electromagnetic radiation falls on a metallic body whose work function is $2 e V$ . For particular radiation of frequency $ν$ , the maximum kinetic energy of the photoelectron is found to be $4 e V$ . Find the maximum kinetic energy (in eV) of photoelectron for the radiation of $\frac{5 ν}{3} .$
$8$
$6$
$16$
$12$

A

16

B

12

C

8

D

6

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Solution

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Electromagnetic radiation falls on a metallic body whose work function is 2eV.
$h ν = ϕ + K E_{m a x} \Rightarrow h ν = 2 + 4$
$\Rightarrow h ν = 6 e V$

Again, $h (\frac{5 ν}{3}) = 2 + K E_{m a x}$

$\frac{5}{3} \times 6 = 2 + K E_{m a x}$

$K E_{m a x} = 8 e V$
So, the answer is option (A).

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