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Question: When an \[\alpha - particle\] of mass m moving with velocity v bombards on a heavy nucleus of charge...

When an αparticle\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:
A. 1m\dfrac{1}{m}
B. 1m\dfrac{1}{{\sqrt m }}
C. 1m2\dfrac{1}{{{m^2}}}
D. m

Explanation

Solution

At distance of closest approaches all the kinetic energy of αparticle\alpha - particle is converted into electrostatic potential energy because at distance of closest approach velocity of αparticle\alpha - particlebecomes zero because of strong repulsion from the heavy nucleus.

Complete step by step solution:
Velocity of αparticle\alpha - particleis v
So initially its kinetic energy [K] will be K=12mv2K = \dfrac{1}{2}m{v^2}
Now assuming αparticle\alpha - particle is bombard from a distance
Then potential energy of αparticle\alpha - particleand heavy nucleus will be approximately zero
Now let's assume at distance of closest approach αparticle\alpha - particlecomes under influence of only one nuclei
Then,
Potential energy at closest distance of approach will be
PE=kq1q2r=k[2e][Ze]roPE = \dfrac{{k{q_1}{q_2}}}{r} = k\dfrac{{[2e][Ze]}}{{{r_o}}}where ro{r_o} is distance of closest approach.
And 2e is charge on αparticle\alpha - particlewhile Ze is charge on the heavy nuclei,
Now applying energy conservation we have,
K=PEK = PE
12mv2=k[2e][Ze]ro\Rightarrow \dfrac{1}{2}m{v^2} = k\dfrac{{[2e][Ze]}}{{{r_o}}}
ro=4kZe2mv2\Rightarrow {r_o} = \dfrac{{4kZ{e^2}}}{{m{v^2}}}
\therefore ro1m{r_o} \propto \dfrac{1}{m}
As we can see that distance of closest approach in αparticle\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 αparticle\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 αparticle\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.