Graphs were plotted between stopping potential ($V_o$) and frequency of incident radiation ($v$) for 2 metals namely Cesium (work fn. 2.14) and Aluminium (work fn. 4.28). The ratio of slope of graph of Cesium to the slope of graph for Aluminium is:

1

The correct answer is Option (3) → 1

According to photoelectric equation -

$eV_o=hv+\phi$

$V_o=\left(\frac{h}{e}\right)v+\phi$   ...(1)

Comparing this with, $y=mx+c$   ...(2)

$∴m=\frac{h}{e}$ = constant

$\frac{m_1}{m_2}=1$