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Light described at a place by the equation $E = (100 V / m)$ $[s i n (5 \times 10^{15} s^{- 1}) t + s i n (8 \times 10^{15} s^{- 1}) t]$ falls on a metal surface having work function $2.0 e V$ . Calculate the maximum kinetic energy of the photoelectrons.

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$1.27 e V$

$2.27 e V$

$3.27 e V$

$4.27 e V$

answer is C.

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Detailed Solution

$f_{h i g h e r} = \frac{8 \times 10^{15}}{2 π} s^{- 1} E = \frac{h f}{1.6 \times 10^{- 19}} e V = \frac{(6.63 \times 10^{- 34}) (8 \times 10^{15})}{1.6 \times 10^{- 19} \times 2 π} = 5.27 e V$
$K_{m a x} = E - W = 3.27 e V$

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