JEE Main · 2004 · 24 Apr 2004 · Offline · Modern Physics

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α

-particle of energy 5 MeV is scattered through

18 0^{\circ}

by a fixed uranium nucleus. The distance of the closest approach is of the order of

(A) $1 \overset{˚}{A}$
(B) $1 0^{- 10} cm$
(C) $1 0^{- 12} cm$ ✓ Correct
(D) $1 0^{- 15} cm$

Answer: (C)

Solution

At closest approach the alpha particle's kinetic energy has turned entirely into electrostatic potential energy,

K E = \frac{1}{4 π ε _{0}} \frac{Z _{1} Z _{2} e ^{2}}{r}

. Putting

Z_{1} = 2

for the alpha particle,

Z_{2} = 92

for uranium and

K E = 5

MeV and solving for

r

gives about

5.3 \times 1 0^{- 14}

m, which is of the order of

1 0^{- 12}

cm.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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