The threshold frequency of a metal with work function 6.63 eV is :

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$16 \times 10^{12} Hz$

$16 \times 10^{15} Hz$

$1.6 \times 10^{12} Hz$

$1.6 \times 10^{15} Hz$

answer is D.

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Detailed Solution

$ϕ_{0} = h v_{0}$
$6.63 \times 1.6 \times 10^{- 19} = 6.63 \times 10^{- 34} v_{0}$
$v_{0} = \frac{1.6 \times 10^{- 19}}{10^{- 34}}$
$v_{0} = 1.6 \times 10^{15} Hz$

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