Question

26. An a-particle is fired towards a nucleus with a momentum p and the
distance of closest approach is r. if its momentums is increased to 4p.
the distance of closest approach will be(एक -कण को एक नाभिक की ओर p
संवेग से प्रक्षेपित किया गया है तथा निकटतम पहुँच दूरी है। यदि इसका संवेग 4p तक
बढ़ा दिया जाय, तो निकटतम पहुँच दूरी होगी)
(a)
(b)
4r
(c) 16r
(d)​

Answer 1

Concept:-In this question we have to use law of conservation of energy...

Solution:-

In that case

kinetic energy of alpha particle during firing=potential energy at the point where it had its maximum approach

We know that,the relation between kinetic energy and momentum is

$kinetic \: energy = \frac{ {p}^{2} }{2m}$

potential energy between two charges q1 and q2

$potential \: energy = \frac{1}{4\pi k} \times \frac{kq1q2}{r}$

using the law of conservation of energy,we can say

$\frac{ {p}^{2} }{2m} = \frac{1}{4\pi k} \times \: \frac{q1q2}{r}$

here everything is constant except r and p

so, we can say

${p}^{2} \: is \: inversely \: proportional\:r$

so,if p become 4 times i.e p^2 becomes 16 times then r will be decreased by 16 times

So ,we can say that the distance of closest approach will be $\frac{r}{16}$

{hope it helps you friend}