1) The work functions of silver and sodium are 4.6 and 2.3 eV, respectively The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is A) 1 B) 4 C) 2 D) .3 Answer: Option AExplanation:$eV_{0}=hf-W$ $\therefore$ $V_{0}=\left(\frac{h}{e}\right)f-\frac{W}{e}$ V0 versus f graph is a straight line with slope= $\frac{h}{e}$= a universal constant. Therefore the ratio of two slopes sholud be 1
