Question 326: Atoms & Nuclei

Question

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

Accepted Answer

Work done to stop the $$\alpha $$ particle is equal to $$K.E.$$
$$	herefore$$ $$qV = {1 \over 2}m{v^2} \Rightarrow q 	imes {{K\left( {Ze} ight)} \over r} = {1 \over 2}m{v^2}$$
$$ \Rightarrow r = {{2\left( {2e} ight)K\left( {Ze} ight)} \over {m{V^2}}} = {{4KZ{e^2}} \over {m{v^2}}}$$
$$ \Rightarrow r \propto {1 \over {{v^2}}}$$ and $$r \propto {1 \over m}.$$

Answer

$${v^2}$$

Answer

$${1 \over m}$$

Answer

$${1 \over {{v^2}}}$$

Answer

$${1 \over {Ze}}$$

Atoms & Nuclei

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

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