The threshold frequency of a metal with work function $6.63 \mathrm{~eV}$ is :

<p>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.</p> <p>The energy of a photon is given by the equation $E = h\nu$, where:</p> <ul> <li>$E$ is the energy of the photon,</li> <li>$h$ is Planck's constant ($h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$), and</li> <li>$\nu$ is the frequency of the photon.</li> </ul> <p>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 $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$:</p> $$ \phi = 6.63 \, \text{eV} \times 1.602 \times 10^{-19} \, \frac{\text{J}}{\text{eV}} = 1.061 \times 10^{-18} \, \text{J} $$ <p>The energy of the photon at the threshold frequency is equal to the work function:</p> $h\nu_0 = \phi$ <p>Solve for $\nu_0$:</p> $$ \nu_0 = \frac{\phi}{h} = \frac{1.061 \times 10^{-18} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} = 1.6 \times 10^{15} \, \text{Hz} $$ <p>Thus, the correct answer is:</p> Option C $1.6 \times 10^{15} \, \text{Hz}$