Solveeit Logo
Login

Question

Physics Question on Atoms

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

A

1 A˚\mathring {A}

B

10−10cm10^{-10}cm

C

10−12cm10^{-12}cm

D

10−15cm10^{-15}cm

Answer

10−12cm10^{-12}cm

Explanation

Solution

According to law of conservation of energy, kinetic energy of α\alpha -particle =potential energy of a-particle at distance of closest approach ie, 12mv2=14πε0q1q2r \frac {1}{2}mv^2= \frac {1}{4 \pi \varepsilon_0} \frac {q_1q_2}{r} ∴5MeV=9×109×(2e)×(92e)r\therefore 5MeV= \frac {9 \times 10^9 \times (2e) \times (92e)}{r} (∵12mv2=5MeV)\bigg (\because \frac {1}{2}mv^2=5MeV \bigg ) ⇒r=9×109×2×92×(1.6×10−19)25×106×1.6×10−19\Rightarrow r=\frac {9 \times 10^9\times2\times92\times(1.6\times10^{-19})^2 }{5\times10^6\times1.6\times10^{-19}} ∴r=5.3×10−14m≈10−12cm\therefore r=5.3 \times 10^{-14}m \approx10^{-12}cm