Question
The threshold frequency of a metal with work function $6.63\ \mathrm{eV}$ is:
✓
Answer
$ \phi_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} \mathrm{~Hz}$
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