# Rutherford experiment [closed]

I am aware that in this experiment originally alpha particles were bombarded on gold foil. But recently I came across a question where this experiment was done with calcium and zinc instead of gold. After a bit of reading I found out that number of alpha particles deflected follows

But then I came across a question asking me to relate it with initial energy of the particle and number of deflecting atoms ( please also explain what exactly this "number of deflecting atoms" refers to)

From my knowledge of electrostatics I know that larger the initial K.E the closer the particle can get to the atom. But I am unsure of what the exact proportion is.

For reference I am also adding a picture of the actual question whose answer is option 1

If possible please give me an exact expression of sorts for the number of alpha particles scattered for a certain scattering angle

P.S. sorry about the poor formatting I was kind of busy. I will reformat the question as soon as possible.

The number of $\alpha$ particle scattered at a particular angle can be determined by finding out the distribution of the scattering angle ($N(\theta) \mathrm d\theta$) where $N(\theta)$ is the number of alpha paricles scattered at angle $\theta$.
First of all, if the alpha particle approach the nucleus with an impact parameter $b$, due to the repulsion the path of the alpha particle will be deflected and it will become an hyperbola, whose eccentricity can be simply calculated from a classical problem of central force motion as, $$\epsilon = \sqrt{1+ \frac{2EL^2}{\mu C^2}}$$ where $E$ is the total energy(which remains conserved and hence initial K.E), $L$ is the angular momentum ( = $\mu v_0b$), here $\mu$ is the reduced mass, and $C$ is $\frac{1}{4\pi \epsilon_0}z_{\alpha}Z^+e^2$.
Now from the knowledge of coordinate geometry, we can say that if the scattering angle is $\theta$, which is actually the supplementary angle between assymptotes of hyperbola $$\sec\left(\frac{\pi}{2}- \frac{\theta}{2}\right)=\epsilon(\text{eccentricity})$$ and also substituting $L^2 = 2\mu Eb^2$ (this is as, $L=\mu v_0b$, and $E=\frac{\mu v_o^2}{2}$ (initial K.E), we will get $$\sqrt{1+\left(\frac{2Eb}C\right)^2}=\frac{1}{\sin(\frac{\theta}{2})}$$ which gives
$$\frac{2Eb}C=\cot(\theta/2)$$ Now, the probablity distribution so that the alpha paticles approach at a particular $$b=P(b)\mathrm db=\rho t2\pi b\mathrm db$$ where $\rho$ is the density of the scattering nuclei and $t$ is the thickness of the foil, but we know from just the previous result that $$b = \frac{C\cot(\theta/2)}{2Eb}$$ now from this if you differentiate the terms to find $b\mathrm db$, and finally calculate the probablity distribution you will get $$P(b)\mathrm db=\frac{\pi}8\rho t\frac{C^2}{E^2}\frac{\sin(\theta)\mathrm d\theta}{\sin^4(\theta/2)}=N(\theta)\mathrm d\theta$$ Now for same scattering angle, the $\theta$ parts will remain constant and for your new case $$C'=\frac{Z_\ce{Ca}}{Z_\ce{Zn}}C=\frac23C$$ and $E'=2E$ , and also $\rho'=8\rho$ (as the density of scattering nuclei in your new case becomes $8$ times of that previous density). So, we will have by making $\theta$ parts constant, $$\frac{N'E'^2}{\rho'C'^2}=\frac{NE^2}{\rho C^2}\to N'=\frac{NE^2\cdot8\rho\cdot\left(\frac23C\right)^2}{\left(2E\right)^2\rho C^2}=\frac89N$$ Thus if $N$ was previously $9000$, it will be now $\frac89\cdot9000=8000$.