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Physics Question on Nuclei

An alpha particle of energy 5 MeV5 \,MeV is scattered through 180∘180^\circ by a fixed uranium nucleus. The distance of closest approach is of the order of

A

1 A˚1\, \mathring{A}

B

10−10cm10^{-10} cm

C

10−12cm10^{-12} cm

D

10−15cm10^{-15} cm

Answer

10−12cm10^{-12} cm

Explanation

Solution

From conservation of mechanical energy decrease in kinetic energy = increase in potential energy or 14πε0(Ze)(Ze)rmin=5 MeV=5×1.6×10−13 J\frac{1}{4 \pi \varepsilon_0} \frac{(Ze)(Ze)}{r_{min}}=5\, MeV=5 \times 1.6 \times10^{-13}\, J

∴   rmin=14πε02Ze5×1.6×10−13\therefore \, \, \, r_{min}=\frac{1}{4 \pi \varepsilon_0}\frac{2Ze}{5 \times 1.6 \times10^{-13}}

15mm=(9×109)(2)(92)(1.6×10−19)25×1.6×10−1315mm = \frac{(9 \times 10^9) (2)(92)(1.6 \times 10^{-19})^2}{5 \times 1.6 \times10^{-13}}

15mm=5.3×10−14 m=5.3×10−12cm15mm = 5.3 \times 10^{-14}\, m= 5.3 \times 10^{-12} cm

i.e.rminr_{min} is of the order of 10−12cm10^{-12} cm