Q 12-11-067JEE MainJEE Main 2024 (27 Jan, Shift 2)Easy
The threshold frequency of a metal with work function $6.63\ \text{eV}$ is:
Answer: (D) $1.6\times10^{15}\ \text{Hz}$
$$\nu_0 = \frac{\phi}{h} = \frac{6.63\times1.6\times10^{-19}}{6.63\times10^{-34}} = 1.6\times10^{15}\ \text{Hz}$$
Solution by Sreeraj P, M.Sc Physics