Sketch of the Graph
(i) Stopping potential and Reason
(ii) Dependence of slope and Explanation
(i) For material B
From the graph for the same value of ' $\mathit{v}$ ', stopping potential is more for material ' $\mathit{B}$ '/ $\left[{\mathit{V}}_0=\frac{\mathit{h}}{\mathit{e}}\left(\mathit{v}-{\mathit{v}}_0\right)\right]$
$\therefore {\mathrm{V}}_0$ is higher for lower value of ${\mathit{v}}_0]$
(ii) No
As slope is given by $\frac{\mathit{h}}{\mathit{e}}$ which is constant.