\(K E_{\max }=E_{v}-\phi\)
or \(2=5-\phi \quad \therefore \phi=3 \mathrm{eV}\)
When \(E_{v}=6 \mathrm{eV}\) then
\(KE_{\max }=6-3=3 \mathrm{eV}\)
or \(e\left(V_{\text {cathode }}-V_{\text {anode }}\right)=3 \mathrm{eV}\)
or \(V_{\text {cathode }}-V_{\text {anode }}=3 \mathrm{V}=-V_{\text {stopping }}\)
\(\therefore V_{\text {stoooing }}=-3 \mathrm{V}\)
(he $=1245 \mathrm{eVnm}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ આપેલ છે.)