\(E_{3}=-13.6 \times \frac{Z^{2}}{n^{2}} \mathrm{eV}=-13.6 \times \frac{4}{9} \mathrm{eV}\)

\(=-13.6 \times \frac{4}{9} \times 1.6 \times 10^{-19} \mathrm{J} \simeq 9.7 \times 10^{-19} \mathrm{J}\)

As per Bohr's model, Kinetic energy of electron in the \(3^{r d}\) orbit \(=-E_{3}\)

\(\therefore \,\,9.7 \times {10^{ - 19}} = \frac{1}{2}{m_e}{v^2}\)

\(v=\sqrt{\frac{2 \times 9.7 \times 10^{-19}}{9.1 \times 10^{-31}}}=1.46 \times 10^{6} \mathrm{ms}^{-1}\)