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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$1 .2 \times 10^{- 18} J$

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

$6 \times 10^{- 12} J$

$1 .2 \times 10^{- 20} J$

answer is C.

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Detailed Solution

$\begin{array}{l} W = h \frac{C}{λ_{0}} \\ W = \frac{6 .6 \times 10^{- 34} \times 3 \times 10^{8}}{330 \times 10^{- 9}} \\ = \frac{6 .6 \times 3 \times 10^{- 18}}{33} \\ = 0 .6 \times 10^{- 18} \\ = 6 \times 10^{- 19} J \end{array}$

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