The distance of 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 closest approach is (1) 4r (2) 2r (3) 2/r (4) r/4

Answer: (4)

All of the kinetic energy of incident particles is converted to the potential energy of the system at the distance of closest approach.

PE = KE

\(\frac{p^{2}}{2m} = \frac{1}{4\pi \epsilon _{0}}\times \frac{q_{1}q_{2}}{r}\)

Initially

\(\frac{p^{2}}{2m}= \frac{2eq}{4\pi \epsilon _{0}r}\)………………………..(1)

Now

\(\frac{2p^{2}}{2m}= \frac{2eq}{4\pi \epsilon _{0}r^{1}}\)………………….(2)

Equation (1) divided by (2)

1/4 = r1/r

r1 = r/4

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