Lost energy of electron is equal to its stopping potential

\(E\, = \,KE\, + \,\Phi \)

\(\frac{12100}{\lambda A}=V+4.47 \mathrm{eV}\)

\(\frac{12100}{2500}=V+4.47\)

\(\frac{124}{25}=V+4.47\)

\(4.95=V+4.47\)

\(\mathrm{V}=4.95-4.47=0.48\)

\(\mathrm{V}_{\text {sphere }}=\frac{K Q}{r}=\frac{9 \times 10^{9} \mathrm{Q}}{100}\)

\(0.48=\frac{9 \times 10^{8} Q}{100}\)

\(Q=\frac{48 \times 10^{9}}{9 \times 100}\)

\(Q=5.5 \times 10^{-13} \mathrm{C}\)