Answer
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Hint: At distance of closest approaches all the kinetic energy of \[\alpha - particle\] is converted into electrostatic potential energy because at distance of closest approach velocity of \[\alpha - particle\]becomes zero because of strong repulsion from the heavy nucleus.
Complete step by step solution:
Velocity of \[\alpha - particle\]is v
So initially its kinetic energy [K] will be \[K = \dfrac{1}{2}m{v^2}\]
Now assuming \[\alpha - particle\] is bombard from a distance
Then potential energy of \[\alpha - particle\]and heavy nucleus will be approximately zero
Now let's assume at distance of closest approach \[\alpha - particle\]comes under influence of only one nuclei
Then,
Potential energy at closest distance of approach will be
\[PE = \dfrac{{k{q_1}{q_2}}}{r} = k\dfrac{{[2e][Ze]}}{{{r_o}}}\]where \[{r_o}\] is distance of closest approach.
And 2e is charge on \[\alpha - particle\]while Ze is charge on the heavy nuclei,
Now applying energy conservation we have,
\[K = PE\]
\[ \Rightarrow \dfrac{1}{2}m{v^2} = k\dfrac{{[2e][Ze]}}{{{r_o}}}\]
\[ \Rightarrow {r_o} = \dfrac{{4kZ{e^2}}}{{m{v^2}}}\]
$\therefore$ \[{r_o} \propto \dfrac{1}{m}\]
As we can see that distance of closest approach in \[\alpha - particle\] bombarding is inversely proportional to its mass.
So, the correct answer is “Option A”.
Note:
Here i have assumed one to one interaction between \[\alpha - particle\] and heavy nuclei but in reality it's not true all other nuclei around the colliding nuclei also contribute in repulsion, still answer will remain same because kinetic energy depends upon mass of \[\alpha - particle\] and inclusion of other nuclei will only change potential energy so we will not get the exactly same expression for distance of closest approach but it will still be inversely proportional to mass as mass is invariable in this context.
Complete step by step solution:
Velocity of \[\alpha - particle\]is v
So initially its kinetic energy [K] will be \[K = \dfrac{1}{2}m{v^2}\]
Now assuming \[\alpha - particle\] is bombard from a distance
Then potential energy of \[\alpha - particle\]and heavy nucleus will be approximately zero
Now let's assume at distance of closest approach \[\alpha - particle\]comes under influence of only one nuclei
Then,
Potential energy at closest distance of approach will be
\[PE = \dfrac{{k{q_1}{q_2}}}{r} = k\dfrac{{[2e][Ze]}}{{{r_o}}}\]where \[{r_o}\] is distance of closest approach.
And 2e is charge on \[\alpha - particle\]while Ze is charge on the heavy nuclei,
Now applying energy conservation we have,
\[K = PE\]
\[ \Rightarrow \dfrac{1}{2}m{v^2} = k\dfrac{{[2e][Ze]}}{{{r_o}}}\]
\[ \Rightarrow {r_o} = \dfrac{{4kZ{e^2}}}{{m{v^2}}}\]
$\therefore$ \[{r_o} \propto \dfrac{1}{m}\]
As we can see that distance of closest approach in \[\alpha - particle\] bombarding is inversely proportional to its mass.
So, the correct answer is “Option A”.
Note:
Here i have assumed one to one interaction between \[\alpha - particle\] and heavy nuclei but in reality it's not true all other nuclei around the colliding nuclei also contribute in repulsion, still answer will remain same because kinetic energy depends upon mass of \[\alpha - particle\] and inclusion of other nuclei will only change potential energy so we will not get the exactly same expression for distance of closest approach but it will still be inversely proportional to mass as mass is invariable in this context.
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