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:
Answer & explanation
Correct answer: option 3
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
$\frac{m_1}{m_2}=1$