In the last lesson we talked about Rutherford's 9 years at McGill University in Canada, during which time he discovered the fickleness of elements and the half-life of radioactive elements, so he won the 1908 Nobel Prize in Chemistry.
Although this award is the greatest honor for scientists, this is not the end of Rutherford's research, he actually always wanted to leave the place where he stayed for 9 years, because Rutherford felt that Canada was a little far from the scientific centers of Europe, especially away from Britain, France and Germany, and it can be said that the most powerful people at that time were in these three places.
In 1906, Rutherford got a professorship at the University of Manchester, and in 1907 Rutherford's laboratory came two young physicists, one was a postdoctoral researcher Hans Geiger, and the other was a university student Marsden.
This is the advantage of returning to the UK, not only the experimental conditions are good, but there are many talents, and the dirty work in the laboratory can be done by young people, and they only need to arrange research projects.
Soon Rutherford decided on the topic of the study, bombarded the gold leaf with α particles, and observed the scattering of α particles after passing through the gold leaf, so there were two reasons for studying this topic:
The first was rutherford's teacher, Thomson, who proposed an atomic model in 1903, in which he said that the positive charge diffuses the space of atoms, and the electrons are embedded in the positive charge, just like the pudding is embedded with raisins. This claim has not been tested experimentally.
The second is that the scattering experiment of α particles itself was done by Rutherford in 1906, when he was still at McGill University, which may be limited to the experimental conditions at the time, and the lack of energy, and did not produce any results.
So his first research in Manchester is this topic, the equipment of this experiment is also simple to say, that is, a α particle source, there was no artificial accelerator at that time, using the natural radioactive element radium released by the α particles, the speed is about 2.09 × 10 ^ 7 m / s, the speed of the α particles can be measured by electromagnetic deflection experiments.
Then we first let the α particles emitted by the radium source pass through a slit-open shield, and after passing through the slit, the α particles will become a narrow bundle, and then we are using this beam of α particles to bombard the gold leaf.
When α particles pass through the gold leaf atom, the α particles will interact with what is inside the gold leaf atom, so that the path of the α particles will be shifted, and then we are letting these scattered α particles shoot onto the back zinc sulfide fluorescent screen, and when the α particle hits the phosphor screen, there will be a flash of light.
So we only need to count the number and position of flashes to know the angular distribution of α particles after being scattered, and of course, we can also see at which angle, α particles are most likely to be scattered. This experiment is easy to say, but in fact, it is a tiring task, which requires counting hours of flashes in an all-black environment. Of course, these tasks were done by Geiger and Marsden.
The first experiments did not find anything special, such as in 1908, Geiger's report to Rutherford, he said, as the deflection angle increased, the number of α particles after scattering would become smaller and smaller, and after a few degrees, α particles could not be seen.
This result was very much in line with expectations at the time, and of course also in line with Thomson's model of the atom, as long as a α particle can penetrate the gold atom, then all the α particles can easily pass through the gold atom.
Because Thomson said that positive charges spread throughout atomic space, such a situation would lead to two possible outcomes: α particles, it would either not pass through one, or all of them would pass;
The result of the experiment was that the positive charge diffuse in the atom as a whole did not affect the α particles, and α particles could pass through the gold atom. The small angled deflection that occurs α particles can be interpreted as interacting with electrons with very small masses.
But in 1909, one day Geiger approached Rutherford and said that Marsden could now do experiments independently, so why not let him do some research as well? Rutherford also thought that Marsden could do it now, so he told Geiger, do you want him to see it, isn't there also a α particles being deflected at a large angle? Rutherford later recalled that he thought it was completely impossible, just a casual remark.
Unexpectedly, two or three days later, Geiger excitedly found Rutherford and told him that not only did he see α particles being deflected out at a large angle, but also that out of the 20,000 α particles, there would be a α particle scattering backwards, which meant that the α particles turned 180 degrees directly like hitting a wall and flew out backwards.
So Rutherford said that, the incredible degree of this phenomenon, like firing a 15-inch shell at a piece of paper, but the shell returned and hit itself.
Because in Rutherford's mind, the mass of α particles is very large, the speed is very high, so the kinetic energy carried is quite considerable, then according to Thomson, there is no such thing in the atom that can directly block α particles. So the results of this experiment are very surprising.
The next key question is how Thomson judged the existence of an atomic nucleus based on the results of this experiment. In fact, this process is very complicated, and it is not so easy for us to hear about it. Because the experimental results were made in 1909, but it was not until 1911 that Rutherford published the concept of the nucleus;
It can be seen that in two years, Rutherford considered many possibilities during this period, solved many ideological difficulties, and finally proposed a method to verify the nuclear model, and only published his own paper after confirming it through experiments. It can also be seen that Rutherford was a very rigorous man and did not like to speculate, which is why he did not like theoretical physicists.
Okay, so let's talk about the process by which Rutherford confirmed the nucleus. First of all, it has to deny its teacher's atomic model, in Geiger and Marsden's experiment, at the angle of 0.87 °, α particles are scattered the most, that is, the probability of being scattered at this angle is the greatest.
But as I said just now, in the 20,000 α particles, there will be such a α particles scattering backwards, indicating that this scattering angle exceeds 90 °, which is the angle between the incidence direction and the exit direction of the α particles, more than 90 degrees, it must be scattered backwards.
This is 100 times larger than the angle of the maximum scattering probability just now, if according to Thomson's model, α particles are interacting with electrons before the deflection occurs, but this deflection angle is very small, and it is impossible to achieve a deflection angle that wants to accumulate to greater than 90 degrees through the action with the electron, which is very small in mathematical probability.
So Thomson speculated that the deflection of the large angle was not the occurrence of multiple collisions, but the deflection of the large angle of the α particle in a collision with something in the atom.
Because the α particles are massive, fast, and positively charged, Rutherford guessed that α particles were likely to hit something of great mass that was also positively charged.
In his 1911 paper, Rutherford calculated a situation in which α particle collides head-on with this positively charged heavy particle, which is like using a leather ball to smash a wall, and the ball's speed drops to 0 in an instant, and the kinetic energy of the ball becomes elastic potential energy, and then the elastic potential energy is converted into the kinetic energy of the ball, and the ball will move in the opposite direction.
Α the collision of the particle and this heavy particle is the same, it also follows the law of conservation of energy, and at first the kinetic energy of the α particle can be calculated according to the square of the mass and velocity of the α particle.
That α particles will feel an electric repulsion when they are close to this positively charged heavy particle, and this electric repulsion force is opposite to the speed direction of the α particle, so the electric repulsion force does a negative work on the α particle, and it can also be said that the kinetic energy carried by the α particle is doing positive work against the electric repulsion force.
In short, when α particle stops near the heavy particle, the work done by the repulsion force is equal to the initial kinetic energy of the α particle, and the work done by the repulse force is equal to (ke×α the charge of the particle × the charge of the heavy particle) / α the distance between the particle and the heavy ion when it is closest.
So we can list a formula based on this relationship, 1/2×α of the mass of the particle×α the initial velocity of the particle² = ke×α the charge of the particle × the charge of the heavy particle) / α the distance when the particle and the heavy ion are closest.
So we can calculate: the distance between the α particle and the heavy particle when it is closest = (2×ke× the charge of the heavy particle) / (the mass-charge ratio of the α particle ×α the initial velocity of the particle²)
In the formula α particle's mass-to-charge ratio and initial velocity are known quantities, but the charge value of the heavy particles is not known, but we can assume that it is a unit charge, that is, the electron charge is worth z times.
So in the end we can calculate that the α particle and the heavy particle collide, the nearest distance is 3.4×z× 10^-16 meters, even if the charge of this heavy particle is hundreds of times the charge of the electron, then this distance is still 1000 times smaller than the size of the gold atom.
So Rutherford speculated that α particles hit something in an atom with a large mass, but a small radius, and a positive charge. Coupled with some of our previous evidence, for example, that the mass of an atom is thousands of times that of an electron, it needs to be explained where the other masses go? Atoms are not charged, but electrons are negatively charged, and where is the positive charge needed to be explained?
Other experiments have also provided references for Rutherford, such as the discovery that cathode rays, or electrons, can travel long distances in gas, which also means that most of the inside of atoms is empty.
So Rutherford envisioned its atomic model, in which the nucleus is positively charged to balance the electron charge and contains the absolute mass of the atom, but the volume is small, and the electrons orbit the nucleus outside the nucleus.
Rutherford needs to verify his ideas before he can publish his paper, and the method of verification is like this, Rutherford needs to calculate according to his own atomic model, what is the probability of α particle being scattered in the range greater than a certain deflection angle, such as what is the probability of α particle being deflected in the range of deflection angle greater than 90 degrees? If the calculated value matches the experimental measurement, then there is no problem with the nucleus.
Well, let's briefly talk about this process, first say a physical quantity called collision parameter, collision parameter is said, α particle has not been deflected by the nucleus, the smallest distance between it and the nucleus is called collision parameter.
It doesn't matter if you don't understand, here I will explain again, α particles are actually not aimed at the nucleus of the atom, but against the gold leaf in the random shooting, so most of the α particles are wiped past the nucleus, just like the above figure, a α particle is now flying forward, it is not facing the nucleus, but has a distance from the nucleus, is misaligned, when it is closest to the nucleus, the distance between it and the nucleus is called the collision parameter.
It can be seen that the smaller the collision parameter, the closer the α particle to the nucleus, the greater the repulsion force, the greater its deflection angle, the greater the collision parameter, the farther the distance, the smaller the repulsion force, and the smaller the deflection angle. The relationship is very clear.
For example, now there is a α particle flying forward, when it passes through the nucleus, the deflection angle is 90 degrees, at this time we can calculate that the collision parameter of this α particle is 1.5 ×z× 10^-16 meters, this z is the same as the previous one, is the number of unit charges of the atomic nucleus. Of course, this calculation process is more complicated, so I will not say it here.
However, we can know that in order to α particles have a deflection angle greater than 90 degrees, the collision parameter must be less than 1.5 × z× 10^-16 meters, that is, α particles must be closer to the nucleus.
Let's calculate what is the probability that α particles are deflected at an angle greater than 90 degrees. A small trick is needed here, that is, to imagine the collision between α particles and the nucleus of an atom as a small disk with the radius of the collision parameters, that is, to imagine the nucleus as a small disk, which is facing the α particles;
For example, when the deflection angle is 90 degrees, the radius of this small disk is 1.5 ×z× 10^-16 meters, if the α particle wants to obtain a greater deflection angle than 90 degrees, it must hit the inside of the disk, of course, the collision parameter is smaller than the radius of this disk. Right, that's understandable.
Therefore, the probability that α particles are scattered to greater than 90 degrees is equal to the proportion of the area of all disks with a radius of 1.5 ×z × 10^-16 meters in gold leaf. That is, the area of each disk is multiplied by the average number of atoms per unit area.
The area formula for the circle is not to be said, let's say that the average number of gold atoms per unit area is calculated, that is, the mass of gold leaf per square meter is divided by the mass of the gold atom.
The mass of gold leaf per square meter is equal to the density of gold multiplied by the thickness of gold leaf, and the mass of gold atoms is also known, so it can be calculated that the number of gold leaf atoms per square meter is 2.3× 10^22.
Multiplying this number by the area of each small disk, the probability of α particles being scattered to greater than 90° in the experiment is 1.6×10^-9z². Geiger's measurement is 1/20000, which is 5×10^-5.
Based on this measurement we can also calculate that the z-value is about 180, which means that the nuclear power charge of the gold atom is 180, which is obviously wrong, and today we know that this value is 79.
However, Rutherford did not use this value in his 1911 paper, but instead used a value measured by a small angle deflection, 97, which is closer. So the probability that Geiger measured in the experiment that α particles are deflected at all angles is basically consistent with the prediction of Rutherford's formula.
This verifies that Rutherford's model of planetary atoms is correct, at least there is nothing wrong with the description of the nucleus. The nuclear power charge of the nucleus is also indirectly measured.
Okay, that's it for today, and we'll talk about accurately measuring the nuclear charge in the next lesson.