Answer
$2.08\times 10^{15}Hz$
Work Step by Step
We can determine the required frequency as follows:
$f=\frac{K_{max}+W_{\circ}}{h}$
We plug in the known values to obtain:
$f=\frac{6.48\times 10^{-19}J+7.328\times 10^{-19}J}{6.63\times 10^{-34}J.s}$
This simplifies to:
$f=2.08\times 10^{15}Hz$
