An alpha nucleus of energy frac{1}{2}mv^2 bom | vedclass

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Question

For closest approach, kinetic energy is converted into potential energy

$\frac{1}{2}m{v^2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{

Vedclass · Accepted Answer

$\frac{1}{m}$

Vedclass · Answer

For closest approach, kinetic energy is converted into potential energy

$\frac{1}{2}m{v^2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r_0}}}$ $ = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{(Ze)(2e)}}{{{r_0}}}$

${r_0} = \frac{{4Z{e^2}}}{{4\pi {\varepsilon _0}m{v^2}}} = $ $\frac{{Z{e^2}}}{{\pi {\varepsilon _0}{v^2}}}\left( {\frac{1}{m}} \right)$

$r_{0}$ is proportional to $\left(\frac{1}{m}\right)$

An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze.$ Then the distance of closest approach for the alpha nucleus will be proportional to

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