Question 602: Atoms & Nuclei

Question

An alpha nucleus of energy $${1 \over 2}$$ mv2  bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

Accepted Answer

Kinetic energy of alpha nucleus is equall
to electrostatic potential energy of the system
of the alpha particle and the heavy nucleus. That
is,
$${1 \over 2}m{v^2} = {1 \over {4\pi {\varepsilon _0}}} 	imes {{\left( {Ze} ight) 	imes \left( {2e} ight)} \over {{d_{in}}}}$$
$$ 	herefore $$ d
min
$$ \propto $$ $${1 \over m}$$

Answer

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

Answer

v2

Answer

$${1 \over m}$$

Answer

$${1 \over {{v_4}}}$$

Atoms & Nuclei

An alpha nucleus of energy ${1 \over 2}$ mv2 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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