In an alpha particle scattering experiment distance of closest approach for the α particle is 4.5 × 10-14 m. If target nucleus has atomic number 80, then maximum velocity of α-particle is _______ × 105 m / s approximately. (1/4 π 0=9 × 109 SI. unit, mass of α particle =.6.72 × 10-27 kg)

JEE Mains · Physics · STD 12 - Atoms

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. \(\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}\right.\) unit, mass of \(\alpha\) particle \(=\) \(\left.6.72 \times 10^{-27} \mathrm{~kg}\right)\)

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Correct Answer

(D) \(158\)

Detailed explanation

\(v=\sqrt{\frac{4 \mathrm{KZe}^2}{\mathrm{mr}_{\min }}}\)

\(=\sqrt{\frac{4 \times 9 \times 10^9 \times 80}{6.72 \times 10^{-27} \times 4.5 \times 10^{-14}}} \times 1.6 \times 10^{-19}\)

\(=9.759 \times 10^{25} \times 1.6 \times 10^{-19}\)

\(=156 \times 10^5 \mathrm{~m} / \mathrm{s}\)

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