Question 555: Atoms & Nuclei

Question

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

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

v
2

Answer

$${1 \over m}$$

Answer

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

Atoms & Nuclei

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