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Q.

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volts is

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a

2

b

4

c

6

d

10

answer is B.

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Detailed Solution

$K_{\max} = (| V_{s} |) e V \Rightarrow | V_{s} | = 4 V$

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