Physics

Physics

Problem

In an alpha particle scattering experiment distance of closest approach for the α\alpha particle is 4.5×10−144.5 \times 10^{-14} m. If target nucleus has atomic number 80, then maximum velocity of α\alpha - particle is ______ ×105\times 10^5 m/s approximately. (14πϵ0=9×109 SI unit, mass of α particle=6.72×10−27 kg)\left(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ SI unit, mass of } \alpha \text{ particle} = 6.72 \times 10^{-27} \text{ kg}\right) ![](![Bookmark icon](https://storage.googleapis.com/diraq-contents/homework-question-parsing/marker/20260106T182719Z/86f6421cf8344f12ac9c807ab7dc12e8/30a26f2d17ca95672702bf50fb4f0242_img.jpg))

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