Light of frequency $\nu$ shining on a metal of work function $\phi_0$ produces photoelectrons with maximum kinetic energy $K_{\max}$. If the incident frequency is now $2\nu$ (with the same metal), the new maximum kinetic energy becomes

1. From Einstein's equation, $K_{\max} = h\nu - \phi_0$. 2. With the new frequency $2\nu$ on the SAME metal, $K'_{\max} = h(2\nu) - \phi_0 = 2h\nu - \phi_0$. 3. Rewrite: $K'_{\max} = (h\nu - \phi_0) + h\nu = K_{\max} + h\nu$. 4. Option A would be right only if $\phi_0 = 0$; option C confuses adding $h\nu$ with adding $\phi_0$; option D would correspond to $\nu/2$, the wrong direction. 5. Note: this is also why a plot of $K_{\max}$ vs $\nu$ is a straight line of slope $h$ — the increment in $K_{\max}$ for any $\Delta\nu$ is exactly $h\,\Delta\nu$. _Source: NCERT Class 12 Physics Part 2, Ch 11, §11.6 (Einstein's equation $K_{\max} = h\nu - \phi_0$), p. 7._