An alpha nucleus of energy 1/2 mv^2 bombards a heavy nuclear target of charge Ze . Then the distance of closest approach for the alpha nucleus will be - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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An alpha nucleus of energy \frac{1}{2} m v^{2} bombards a heavy nuclear target of charge Z e . Then the distance of closest approach for the alpha nucleus will be proportional to:

Solution & Step-by-step Explanation

At the distance of closest approach r_{0}, the initial kinetic energy is completely converted into electrostatic potential energy. K E = PE ⟹ \frac{1}{2} m v^{2} = \frac{1}{4 π ϵ _{0}} \frac{( 2 e ) ( Z e )}{r _{0}} Rearranging for r_{0} : r_{0} = \frac{1}{4 π ϵ _{0}} \frac{4 Z e ^{2}}{m v ^{2}} From this expression: r_{0} \propto \frac{1}{m} and r_{0} \propto \frac{1}{v ^{2}} .

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An alpha nucleus of energy \frac{1}{2} m v^{2} bombards a heavy nuclear target of charge Z e . Then the distance of closest approach for the alpha nucleus will be proportional to:

\frac{1}{Z e}

v^{2}

\frac{1}{m}

\frac{1}{v ^{4}}

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