The threshold frequency, $$ \nu_0 $$, corresponds to the minimum frequency of light required to eject electrons from the surface of a metal, a phenomenon known as the photoelectric effect. The work function, represented by $$ \phi $$, is the minimum energy needed to remove an electron from the surface of the metal.

The energy of a photon is given by the equation $$ E = h\nu $$, where:

• $$ E $$ is the energy of the photon,
• $$ h $$ is Planck's constant ($$ h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} $$), and
• $$ \nu $$ is the frequency of the photon.

To find the threshold frequency for a metal with a work function of $$ 6.63 \, \text{eV} $$, we must first express the work function in joules (since Planck's constant is in joules per second). To convert electron volts to joules, use the conversion factor $$ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} $$:

$$ \phi = 6.63 \, \text{eV} \times 1.602 \times 10^{-19} \, \frac{\text{J}}{\text{eV}} = 1.061 \times 10^{-18} \, \text{J} $$

The energy of the photon at the threshold frequency is equal to the work function:

$$ h\nu_0 = \phi $$

Solve for $$ \nu_0 $$:

$$ \nu_0 = \frac{\phi}{h} = \frac{1.061 \times 10^{-18} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} = 1.6 \times 10^{15} \, \text{Hz} $$

Thus, the correct answer is:

Option C $$ 1.6 \times 10^{15} \, \text{Hz} $$