Q30. In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10 ^ { - 14 } \mathrm {~m}$. If target nucleus has atomic number 80 , then maximum velocity of $\alpha$ - particle is $\_\_\_\_$ $\times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }$ approximately. $\left( \frac { 1 } { 4 \pi \epsilon _ { 0 } } = 9 \times 10 ^ { 9 } \right.$ SI unit, mass of $\alpha$ particle $\left. = 6.72 \times 10 ^ { - 27 } \mathrm {~kg} \right)$