Physics · Atomic Physics

JEE Main 2024 — 8 April, Shift 1 — Question 57

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 $\qquad$ $\times 10^{5}$ $\mathrm{m} / \mathrm{s}$ approximately. $\left(\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9}\right.$ SI unit, mass of $\alpha$ particle $=$ $\left.6.72 \times 10^{-27} \mathrm{~kg}\right)$

Answer: 156

Numerical answer — enter this value.

Step-by-step solution

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

$\begin{array}{*{35}{r}}{} & ~=\sqrt{\frac{4\times 9\times {{10}^{9}}\times 80}{6.72\times {{10}^{-27}}\times 4.5\times {{10}^{-14}}}}\times 1.6\times {{10}^{-19}} \\{} & ~=9.759\times {{10}^{25}}\times 1.6\times {{10}^{-19}} \\{}&~=156\times {{10}^{5}}\text{ }\!\!~\!\!\text{ m}/\text{s} \\\end{array}$

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Physics
Chapter: Atomic Physics
Topic: Rutherford's and Bohr's Model of Atom

Answer: 156

Step-by-step solution

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