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Q.

If threshold wavelength $(λ^{0})$ for ejection of electron from metal is 330 nm, then work function for the photoelectric emission is

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a

$6 \times 10^{- 10} J$

b

$1.2 \times 10^{- 18} J$

c

$3 \times 10^{- 19} J$

d

$6 \times 10^{- 19} J$

answer is D.

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Detailed Solution

$W_{0} = \frac{hc}{λ_{0}} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8} J}{330 \times 10^{- 9}}$

$= 6 \times 10^{- 19} J$

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