Question 566: Atoms & Nuclei

Question

When an $$\alpha $$-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as

Accepted Answer

At closest distance of approach, the kinetic
energy of the particle will convert completely
into electrostatic potential energy.
Kinetic energy = $${1 \over 2}m{v^2}$$
Potential energy = $${{KQq} \over r}$$
$$ 	herefore $$ $${1 \over 2}m{v^2}$$ = $${{KQq} \over r}$$
$$ \Rightarrow $$ r $$ \propto $$ $${1 \over m}$$

Answer

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

Answer

m

Answer

$${1 \over m}$$

Answer

$${1 \over {\sqrt m }}$$

Atoms & Nuclei

When an $\alpha $-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as

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