Question

The distance of closest approach of an $\alpha $ particle fired towards a nucleus with momentum p is r. The distance of closest approach when the momentum of $\alpha $ is 2p is-
A. 2r
B. 4r
C. r/2
D. r/4

Answer 1

Hint- We will equate the values of potential energy and kinetic energy in the solution. Then we will put the formulas equal by writing the formula of kinetic energy in terms of momentum because momentum is given in the question.
Formula used: $K.E = \dfrac{1}{2}.\dfrac{{{m^2}{v^2}}}{m}$ and $P.E = \dfrac{1}{{4\pi {\varepsilon _0}}}.\dfrac{{{q_1}{q_2}}}{r}$

Complete Step-by-Step solution:
Since we are talking about the closest approach in the question, the potential energy at the time will be equal to initial ${K_{\max }}$.
$ \Rightarrow P.E = K.E$
Momentum (p) is given in the question so we will write kinetic energy in terms of momentum, we get-
$ \Rightarrow K.E = \dfrac{1}{2}.\dfrac{{{m^2}{v^2}}}{m}$
Since we know that momentum p is equal to mv, $p = mv$, the above formula becomes-
$ \Rightarrow K.E = \dfrac{1}{2}.\dfrac{{{p^2}}}{m}$
As we already knew-
$
   \Rightarrow P.E = K.E \\
    \\
   \Rightarrow P.E = \dfrac{{{p^2}}}{{2m}} \\
$
The formula of potential energy is-
$ \Rightarrow P.E = \dfrac{1}{{4\pi {\varepsilon _0}}}.\dfrac{{{q_1}{q_2}}}{r}$ (where r is the closest approach)
So, from this we can say that-
$
   \Rightarrow P.E = \dfrac{{{p^2}}}{{2m}} \\
    \\
   \Rightarrow \dfrac{1}{{4\pi {\varepsilon _0}}}.\dfrac{{{q_1}{q_2}}}{r} = \dfrac{{{p^2}}}{{2m}} \\
$
Thus, from the above equation it is clear that r is inversely proportional to ${p^2}$.
So, if we double the value of momentum, the value of r will become $\dfrac{r}{4}$.
This clearly states that option D is the correct option.

Note: The kinetic energy (KE) of a particle is the energy that it has because of its movement. It is characterized as the work needed to accelerate a body of a given mass from rest to its expressed speed.