00:01
Here we have a problem from modern physics and the topic is rather for scattering.
00:14
In this experiment we sent some alpha particles with certain energy, let's call it e sub alpha.
00:27
You can get alpha particles.
00:30
Towards some target and the target here is just some aluminum.
00:38
Foil having certain thickness t and what we observe is that these alpha particles will emerge with some angle theta after scattering through these aluminum atoms in the foil and we are asked to find the fraction of particles that are scattered through some angles let us do that and let us start at the formula so we have this fraction of part of the particles that are scattered at angles larger than some theta given by f sub theta equals the cross section of this interaction of this interaction times the column density not meet and the expressions are pi times b squared times t now here b is the impact parameter and n is the number of target atoms per volume so it is a material property having nothing to do with this experiment so it is just an independent quantity by itself now let us get the column density of this aluminum foil having some thickness t equals two micrometers of aluminum with thickness two equals two equals two equals two micrometers.
02:57
Nt equals so what is this n value, namely the number of type atoms per volume? it is the density of the aluminum particles or aluminum atoms basically times the avocado number divided by the molar mass of the aluminum.
03:25
And we have this t right next to it.
03:27
So what are these numerical values of these numbers? so we have to do some googling to find that the aluminum has a density given by 2 .7 grams per centimeter cubed.
03:47
And the avogadro number is given by 6 .02 times 10 to the power 23.
03:58
Particles per 1 mole and the molar mass of the aluminum is 26 .982 grams per mole and we have this thickness equals 2 necometers now i'm not going to insert the numbers here i'm going to do it at the end so let us just keep going the calculations.
04:33
Okay, we have the column density, we have one half of the fraction formula and in the second half let us look at the cross section which is given by pi times the square of the impact parameter.
04:51
So we have pi times, okay the impact parameter is given as follows.
04:56
The coulomb constant is equal to 1 over 4 pi x0 times the electric charge of alpha particles times the electric charge of the nucleus of this aluminum atoms divided by two times the energy of the alpha particles times the contangent of half of this scattering angle that's it now let us see what are the numerical values we have the coulomb constant given by 8 .99 times 10 to the power 9.
05:44
Newton per inter squared divided by coulomb squared.
05:50
The electric charge of alpha particles is just plus 2e and the sign will not matter here because we are taking the square.
06:00
So it is just 2e.
06:02
The charge of this aluminum nuclei is given by the atomic number of aluminum times the electron charge so it's again positive so when we look at the periodic table we see that the aluminum has the atomic number 13 we are given the energy of the alpha particles to be 7 .0 m .ev and we know that the electron charges is equal to 1 .6 times 10 to the power 19 cool -on and one electron volt corresponds to an energy of 1 .6 times 10 to the power minus 19 joules.
06:57
So now we can insert everything into our calculators except this co -tangent factor then i get 2 .0 0 .7909 5 times 10 to the power minus 6 times contention of this half of the scattering angles so i am keeping some decimal some figures after a decimal point just not to lose any accurate or precision.
07:38
So this is the fraction of alpha particles that are scattered through some angle greater than theta...