What is the work function of the metal if the light of wave length $4000 A^{0}$ generates photo electrons of velocity $6 \times 10^{5} {ms}^{- 1}$
(mass of $e^{-} = 9 \times 10^{- 31},$ $v = 3 \times 10^{8} {ms}^{- 1}, h = 6 .6 \times 10^{- 34} J 1 ev = 1 .602 \times 10^{- 19} J)$

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3.1 ev

2.1 ev

0.9 ev

4.0 ev

answer is C.

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Detailed Solution

$\begin{array}{l} K . E = hv - ω_{o} (hv = \frac{c}{λ}) \\ ω_{o} = hv - K . E (K . E = \frac{1}{2} {mv}^{2}) \\ ω_{o} = \frac{6 .6 \times 10^{- 34} \times 3 \times 10^{8}}{4 \times 10^{3} \times 10^{- 10}} - \frac{1}{2} \times 9 \times 10^{- 31} {(6 \times 10^{5})}^{2} \\ = 5 \times 10^{- 9} - 1 .62 \times 10^{- 19} = 3 .38 \times 10^{- 19} J \\ 1 e v = 1 .602 \times 10^{- 19} J \\ ? = 3 .38 \times 10^{- 19} J \\ = 2 .1 e v \end{array}$