Question: Q. 2. (i) Draw a schematic arrangement of GeigerMarsden experiment showing the scattering of $\alpha$-particles by a thin foil of gold. Why is it that most of the $\alpha$-particles go right through the foil and only a small fraction gets scattered at large angles ? Draw the trajectory of the $\alpha$-particle in the coulomb field of a nucleus. What is the significance of impact parameter and what information can be obtained regarding the size of the nucleus ?
(ii) Estimate the distance of closest approach to the nucleus $(Z=80)$ if a $7.7 \mathrm{MeV} \alpha$-particle before it comes momentarily to rest and reverses its direction.
U A [Delhi I, II, III 2015]
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Solution:
Ans. (i)
For most of the $\alpha$-particles, impact parameter is large, hence they suffer very small repulsion due to nucleus and go right through the foil.
It gives an estimate of the size of nucleus.
1
(ii) K.E of the $\alpha$-particle $=$ potential energy possessed by beam at distance of closest approach.
$$ \begin{aligned} \frac{1}{2} m v^{2} & =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{(2 e)(Z e)}{r_{0}} \ 7.7 \times 1.6 \times 10^{-3} & =\frac{9 \times 10^{9} \times 2 \times 2.56 \times 10^{-38} \times 80}{r_{0}} \quad 1 / 2 \ r_{0} & =\frac{9 \times 10^{9} \times 2 \times 2.56 \times 10^{-38} \times 80}{7.7 \times 1.6 \times 10^{-13}} \mathrm{~m}^{1 / 2} \ & =299 \times 10^{-16} \mathrm{~m} \ & =29.9 \times 10^{-15} \mathrm{~m} \approx 30 \times 10^{-15} \mathrm{~m}^{1 / 2} \end{aligned} $$
