Appeared in: 2026 (2 sets)
(IV) The threshold frequency for a metal surface is $\nu_0$. If the radiation of frequency $3\nu_0$ illuminates the surface, the maximum kinetic energy (KE) of photoelectrons is $E_1$. If the frequency were increased to $6\nu_0$, the maximum KE of the photoelectrons becomes $E_2$. Then $\left(\frac{E_1}{E_2}\right)$ equals
(A) $1/3$
(B) $1/2$
(C) $2/5$
(D) $3/4$
OR
Let m be the slope of the graph line for metal B. If e is the value of electron charge, then Planck's constant 'h' is given by
(A) $me$
(B) $\frac{1}{me}$
(C) $\frac{m}{e}$
(D) $\frac{e}{m}$