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

The threshold frequency is given by:

$$\phi = h\nu_0$$

$$\nu_0 = \frac{\phi}{h}$$

$$\phi = 6.63$$ eV $$= 6.63 \times 1.6 \times 10^{-19} = 10.608 \times 10^{-19}$$ J

$$\nu_0 = \frac{10.608 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.6 \times 10^{15}$$ Hz

The answer is $$1.6 \times 10^{15}$$ Hz, which corresponds to Option (4).

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