QJEEM24Ph46bc: JEE Main 2024 Physics - Class 12 JEEM24Ph46bc Solution: The threshold frequency of a metal with work function is :

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 $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$ :

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

Solve for $\nu_0$ :

Thus, the correct answer is: