Question

In an alpha particle scattering experiment, distance of closest approach for the \(\alpha\) particle is \(4.5 \times 10^{-14} \, \text{m}\). If the target nucleus has atomic number 80, then maximum velocity of \(\alpha\)-particle is ______ \( \times 10^5 \, \text{m/s} \) approximately.
\[ \left( \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{SI unit}, \, \text{mass of } \alpha \, \text{particle} = 6.72 \times 10^{-27} \, \text{kg} \right) \]

Answer 1

Correct Answer: 156

The distance of closest approach is given by:
\[r_{\text{min}} = \frac{4KZe^2}{mv^2}.\]
Rearranging for velocity:
\[v = \sqrt{\frac{4KZe^2}{mr_{\text{min}}}}.\]
Substitute values:
\[v = \sqrt{\frac{4 \cdot 9 \cdot 10^9 \cdot 80 \cdot (1.6 \times 10^{-19})^2}{6.72 \times 10^{-27} \cdot 4.5 \times 10^{-14}}}.\]
Simplify:
\[v = 156 \times 10^5 \, \text{m/s}.\]
Final Answer:
$156 \times 10^5 \, \text{m/s}$.