Question

The distance of the closest approach of an alpha-particle fired towards a nucleus with momentum p is r. If the momentum of the alpha-particle is 2p, the corresponding distance of the closest approach is:
$$\begin{gathered} (a)\text{ 4r} \\ (b)\text{ 2r} \\ (c){\displaystyle \frac{2}{r}} \\ (d){\displaystyle \frac{r}{4}} \end{gathered}$$

Answer 1

Hint: In this question use the relationship between the distance of the closest approach, Kinetic energy, atomic number and the charge on particle that is $r={\displaystyle \frac{1}{4\pi {\xi }_o}}{\displaystyle \frac{2Z{e}^2}{\left(K.E\right)}}$, then use the relationship of the kinetic energy in terms of momentum that is $K.E={\displaystyle \frac{1}{2}}{\displaystyle \frac{p^2}{m}}$, this helps establishing relationship between the distance of the closest approach and the momentum. This concept will help approach the solution of this problem.

Complete step-by-step answer:
The distance of the closest approach of an alpha particle fired towards a nucleus is given as
$r={\displaystyle \frac{1}{4\pi {\xi }_o}}{\displaystyle \frac{2Z{e}^2}{\left(K.E\right)}}$.............. (1)
Where, Z = atomic number of the nuclei, e = charge on particle = $1.6\times {10}^{-19}eV$, and K.E = kinetic energy of the particle.
Now as we know that the kinetic energy of the particle is given as
$\Rightarrow K.E={\displaystyle \frac{1}{2}}m.v^2$................... (2)
Where, m = mass of the particle
              v = velocity of the particle
Now as we all know that the momentum is the product of the mass of the particle and the velocity of the particle so we have,
$P=mv$................. (3)
Where, P = momentum
             m = mass of the particle
             v = velocity of the particle
Now from equation (2) and (3) we have,
$\Rightarrow K.E={\displaystyle \frac{1}{2}}{\displaystyle \frac{p^2}{m}}$..................... (4)
Now from equation (1) and (4) we have,
$r={\displaystyle \frac{1}{4\pi {\xi }_o}}{\displaystyle \frac{2Z{e}^2}{\left({\displaystyle \frac{1}{2}}{\displaystyle \frac{p^2}{m}}\right)}}$
Therefore, the distance (r) of the closest approach of an alpha particle fired towards a nucleus is inversely proportional to the square of the momentum so we have,
$\Rightarrow r\propto {\displaystyle \frac{1}{P^2}}$.................. (5)
Now it is given that if the momentum of the alpha-particle is 2P (i.e. $P_1=2P$) so the corresponding distance of the closest approach is,
Let the corresponding distance of the closest approach be $r_1$
Therefore from equation (5) we have,
${\displaystyle \frac{r_1}{r}}={\displaystyle \frac{P^2}{P_1^2}}$
Now substitute the value we have,
$\Rightarrow {\displaystyle \frac{r_1}{r}}={\displaystyle \frac{P^2}{{\left(2P\right)}^2}}={\displaystyle \frac{1}{4}}$
$\Rightarrow r_1={\displaystyle \frac{r}{4}}$
So this is the required corresponding distance of the closest approach.
So this is the required answer.
Hence option (D) is the correct answer.

Note – The main reason behind gaining of the kinetic energy is that when an object works or speeds up then some net force is applied by the object onto the ground, this helps in production of energy when we talk about translational motion. An alpha particle in general contains two protons and two neutrons. Alpha particles are produced as a result of alpha decay.