Answer
$$8.69 \times 10^{6} \mathrm{\ m} / \mathrm{s} $$
Work Step by Step
The initial speed of the neodymium nucleus is
$$v_{\mathrm{Nd}}=\sqrt{\frac{2 K_{\mathrm{Nd}}}{M_{\mathrm{ND}}}}=\sqrt{\frac{2\left(60 \times 10^{6} \mathrm{eV}\right)\left(1.60 \times 10^{-19} \mathrm{J} / \mathrm{eV}\right)}{(153 \mathrm{u})\left(1.661 \times 10^{-27} \mathrm{kg} / \mathrm{u}\right)}}=8.69 \times 10^{6} \mathrm{\ m} / \mathrm{s} $$
