If the work functions of three photosensitive materials are $1 e V$ , $2 e V$ and $3 e V$ respectively, then the ratio of the respective frequencies of light that produce photoelectrons of maximum kinetic energy of $1 e$ V from each of them is:

1 : 2 : 3

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2 : 3 : 4

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1 : 1 : 1

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3 : 2 : 1

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4 : 3 : 2

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Solution

The correct option is B $2 : 3 : 4$
From Einstein photo electric equation
$h v = K E + ϕ$
where, $ϕ =$ work functione $= h v_{0}$
$K E = 1$ eV
$h v_{1} = 1 + 1 = 2$ eV
$h v_{2} = 1 + 2 = 3$ eV
$h v_{3} = 1 + 3 = 4$ eV
$\Rightarrow v_{1} : v_{2} : v_{3} = 2 : 3 : 4$ .

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