In an alpha particle scattering experiment distance of closest approach for the \alpha particle is 4.5 \times 10^{-14} {\m}. If target nucleus has atomic number 80 , then maximum velocity| of \alpha-particle is ______ \times 10^5 {\m} \/ { s} approximately. (\frac{1}{4 \pi E_0}=9 \times 10^9 {\SI}. unit, mass of \alpha particle = 6.72 \times 10^{-27} {\kg}) [8 Apr 2024 Shift 1]

Atoms and Nuclei Part 3

Section: Physics

Question : 41 of 87

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In an alpha particle scattering experiment distance of closest approach for the α particle is 4.5×10−14m. If target nucleus has atomic number 80 , then maximum velocity| of α-particle is ______ ×105 m∕ s approximately.
(‌

4πE0

=9×109‌SI. unit, mass of α particle = 6.72×10−27‌kg)

[8 Apr 2024 Shift 1]

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