Question

The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. Find out the ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium.

Answer 1

We have the relation $he=h{v}_o+KE$

=$h{v}_o+e{V}_o$ [$V_o$ =stopping potential ]

So $V_o=h/e(v-v_o)$

Or $V_o=\left(h/e\right)v-\left(h/e\right)v_o$

This is of the form $y=mx+c$

Slope =$m=h/e$ and y-intercept =$-h{v}_o/e$

Slope of this graph is $h/e$. So the slope does not depend on work function. So the ratio is 1.