Photoelectric emission from a metal stops below a certain frequency of incident light, called the threshold frequency $\nu_0$. The threshold frequency depends primarily on which factor?

1. Einstein's photoelectric equation is $h\nu = \phi_0 + K_{\max}$. 2. The threshold corresponds to $K_{\max} = 0$, giving $h\nu_0 = \phi_0$, or $\nu_0 = \phi_0 / h$. 3. Since $h$ is a universal constant, $\nu_0$ is determined solely by $\phi_0$, which is a property of the metal. 4. Intensity controls how many photoelectrons are emitted per second (option A wrong); plate area and temperature do not change the per-photon emission threshold (options B, D wrong). _Source: NCERT Class 12 Physics Part 2, Ch 11 "Dual Nature of Radiation and Matter", §11.4.3 + §11.6, p. 5–7._