Atoms & Nuclei

JEE Main level, numerical answer, medium

10 of 239

Question 10 of 239

In an alpha particle scattering experiment distance of closest approach for the α\alpha particle is 4.5×10−14 m4.5 \times 10^{-14} \mathrm{~m}. If target nucleus has atomic number 80 , then maximum velocity of α\alpha - particle is _______ ×105 m/s\times 10^5 \mathrm{~m} / \mathrm{s} approximately.
(14πϵ0=9×109\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9\right. SI unit, mass of α\alpha particle =6.72×10−27 kg)\left.=6.72 \times 10^{-27} \mathrm{~kg}\right)

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Answer: 156

v=4KZe2mrmin =4×9×109×806.72×10−27×4.5×10−14×1.6×10−19=9.759×1025×1.6×10−19=156×105 m/s\begin{aligned} v & =\sqrt{\frac{4 \mathrm{KZe}^2}{\mathrm{mr}_{\text {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}\end{aligned}

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