JEE MAIN - Physics (2006 - No. 17)
An alpha nucleus of energy $${1 \over 2}m{v^2}$$ bombards a heavy nuclear target of charge $$Ze$$. Then the distance of closest approach for the alpha nucleus will be proportional to
$${v^2}$$
$${1 \over m}$$
$${1 \over {{v^2}}}$$
$${1 \over {Ze}}$$
Explanation
Work done to stop the $$\alpha $$ particle is equal to $$K.E.$$
$$\therefore$$ $$qV = {1 \over 2}m{v^2} \Rightarrow q \times {{K\left( {Ze} \right)} \over r} = {1 \over 2}m{v^2}$$
$$ \Rightarrow r = {{2\left( {2e} \right)K\left( {Ze} \right)} \over {m{V^2}}} = {{4KZ{e^2}} \over {m{v^2}}}$$
$$ \Rightarrow r \propto {1 \over {{v^2}}}$$ and $$r \propto {1 \over m}.$$
$$\therefore$$ $$qV = {1 \over 2}m{v^2} \Rightarrow q \times {{K\left( {Ze} \right)} \over r} = {1 \over 2}m{v^2}$$
$$ \Rightarrow r = {{2\left( {2e} \right)K\left( {Ze} \right)} \over {m{V^2}}} = {{4KZ{e^2}} \over {m{v^2}}}$$
$$ \Rightarrow r \propto {1 \over {{v^2}}}$$ and $$r \propto {1 \over m}.$$
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