Question

When a silver foil (Z=47) was used in an α-ray scattering experiment, the number of α-particles scattered at 30 was found to be 200 per minute. If the silver foil is replaced by aluminum (Z=13) foil of the same thickness, the number of α-particles scattered per minute at 30 is nearly equal to
A. 15
B. 30
C. 10
D. 26

Answer 1

As we all know clearly that Rutherford did the gold foil experiment to understand the nature of the atom. In his experiment, he concluded there is a dense center inside the atom called the nucleus which is positively charged. In his experiment, he found that the number of alpha particles scattered per unit time is directly proportional to the atomic number ($Z_e$) of the foil element.

Complete step by step answer:
As we all know that the number of alpha particles scattered by a foil is given by the relation:
$N \propto \dfrac{{{Z_e}^2}}{{{{\sin }^4}\left( {\dfrac{\theta }{2}} \right)}}$
Here N is the number of alpha particles scattered per minute, Ze is the atomic number of the foil element used, and $\theta$ is the scattering angle.
So, for the silver foil, we can write the relation as,
$\Rightarrow {N_s} \propto \dfrac{{{Z_s}^2}}{{{{\sin }^4}\left( {\dfrac{\theta }{2}} \right)}}$ ……. (I)
Here $N_s$ is the number of alpha particles scattered per minute in silver foil, $Z_s$ is the atomic number of the silver foil element used, and $\theta$ is the scattering angle.
Again, for the silver foil, we can write the relation as,
$\Rightarrow {N_{al}} \propto \dfrac{{{Z_{al}}^2}}{{{{\sin }^4}\left( {\dfrac{\theta }{2}} \right)}}$ …….(II)
Here Nal is the number of alpha particles scattered per minute in aluminum foil, $Z_{al}$ is the atomic number of the aluminum foil element used, and $\theta$ is the scattering angle.
So we can divide equation (I) with equation (II) to get the useful relation. So, we will get,
$\Rightarrow \dfrac{{{N_s}}}{{{N_{al}}}} \propto \dfrac{{\dfrac{{{Z_s}^2}}{{{{\sin }^4}\left( {\dfrac{\theta }{2}} \right)}}}}{{\dfrac{{{Z_{al}}^2}}{{{{\sin }^4}\left( {\dfrac{\theta }{2}} \right)}}}}$
Since the scattering angle is the same so, after simplifying it further the scattering angle term will cancel out and we will get,
$\Rightarrow \dfrac{{{N_s}}}{{{N_{al}}}} = \dfrac{{{Z_s}^2}}{{{Z_{al}}^2}}$…… (III)
Now we will substitute $N_s=200$, $Z_{al} = 13$, and $Z_s=47$ in equation (III) to find the value of Nal.
So, we get:
$\Rightarrow \dfrac{{200}}{{{N_{al}}}} = \dfrac{{{{47}^2}}}{{{{13}^2}}}$
$\Rightarrow {N_{al}} = \dfrac{{200 \times {{13}^2}}}{{{{47}^2}}}$
$\therefore {N_{al}} \approx 15$

Therefore, the number of alpha particles scattered per min for aluminium foil is 15 and the correct option is A.

Additional information:
There is a necessary condition for the minimum deviation, the prism lies symmetrically with respect to the incident ray and the emergent ray i.e.($i=e$). The angle between the emergent ray and the direction of the incident ray is called the angle of deviation. There is one and only one angle of incidence for which the deviation produced by the prism is minimum. For any other angle of deviation, there are two values of ($i$) and ($e$).

Note:
We should keep in mind that Rutherford in his experiment never told about neutrons but also said that the nucleus also is empty spaced and since the alpha particles are deflecting then the nucleus is composed of positive charges called protons. The alpha particle is generally $He^{2+}$ ions with no electrons in it which are emitted by radioactive material.