Question

What is the distance of closest approach to the nucleus of an $$\alpha$$ -particle which
undergoes scattering by 180° in the Geiger-Marsden experiment?

Answer 1

Hint: In the year 1909, this experiment was performed by Hans Geiger and Ernest Marsden, under the direction of Ernest Rutherford. In this case, when the particle reaches the closest distance to the nucleus, it will come to rest and its initial kinetic energy will be completely transformed into potential energy.

Complete step by step solution: In this case, when the $$\alpha$$ particle reaches to the closest distance to the nucleus, it will come to rest and its initial kinetic energy will be completely transformed into potential energy.

This condition can be given by the relation,
$${\displaystyle \frac{1}{2}}m{v_i}^2={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{(2e)(Ze)}{r_0}}$$
where, v = initial speed
Z = atomic number of the nucleus
m = mass of the alpha-particle
$$r_0$$= distance of closest approach of the $$\alpha$$ particle to the nucleus.
e = charge of an electron = $$1.6\times {10}^{-19}$$
For the Geiger-Marsden experiment’
$$\begin{array}{rl}& {\displaystyle \frac{1}{2}}m{v}_i=5.5MeV\\ & =5.5\times {10}^6\times 1.6\times {10}^{-19}\\ & =8.8\times {10}^{-13}\end{array}$$
Now substituting the value to kinetic energy in the above equation,
$$8.8\times {10}^{-13}={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{(2e)(Ze)}{r_0}}$$
$$\begin{array}{rl}& r_0=4.136\times {10}^{-15}\\ & r_0=4.3fm\end{array}$$
After rearranging,
$$r_0={\displaystyle \frac{1}{4\times 3.14\times 8.86\times {10}^{-12}}}{\displaystyle \frac{2\times 79\times {(1.6\times {10}^{-19})}^2}{8.8\times {10}^{-13}}}$$
$$\begin{array}{rl}& r_0=412.82\times {10}^{-16}\\ & r_0=41.2fm\end{array}$$
Therefore, the distance of closest approach is $$41.2fm$$.


Note: In Rutherford’s famous $$\alpha$$ particle scattering experiments, $$\alpha$$ particle ($$+2e=2\times 1.6\times {10}^{-19}C$$, mass = $$6.6\times {10}^{-27}$$kg) were fired at a gold nucleus ($$+79e=+79\times 1.6\times {10}^{-19}C$$). We have used the values accordingly. Energy of the alpha particle is also taken as 5.5 MeV.