The work functions of two metals X and Y are 2.5eV and 4.5 eV respectively. The ratio of slope of stopping potential versus frequency plot for Y to that of X is

$\begin{array}{l}hv=h{v}_0+\frac{1}{2}m{v}_{max}^2\\ \frac{hc}{\lambda }=h{v}_0+e{V}_s\\ v_s=\left(\frac{hc}{e}\right)\frac{1}{\lambda }-\frac{h{v}_0}{e}\\ y=mx+c\\ m=\frac{hc}{e}\end{array}$

The slope of stopping potential versus frequency plot is independent of metal and is constant. Therefore, the ratio is 1 : 1.