Rutherford Alfa experiment and observations about Atom

Physics is a branch of science where we study nature. It deals with different properties of matter. We know that the matter has three different states like solids state, liquid state and gaseous state. All the states the matter consists of molecules. It is understood that molecules can be divided into atoms. When they were named as atoms, it means they are not divisible. But later we are able to divide atoms further into nucleus and the electrons which are revolving around the nucleus.

To know the structure of atoms, we have different theories and the corresponding experimental results.

We are having a model called pulp pudding model. It is also called watermelon model. According to this theory electrons and protons are uniformly distributed in the nucleus like the seeds of a watermelon. As this theory is unable to explain all the properties of matter, a new theory called Rutherford Alfa experimental model is introduced.

According to Rutherford Alfa experiment model nucleus is consisting of positive charges and the size of the nuclei is much smaller than that of the atom. To verify this, an experiment is performed. In this experiment helium particles are allowed to strike a thin gold foil. Helium particles consist of positive charge. As the gold foil is very thin, we can assume that helium particles can directly go and interact with the atom itself.

It is experimentally observed that, out of 8000 alfa rays that were driven towards the nucleus, only one is deviated. From that it can be concluded that the nucleus is also having a positive charge. Then only there will be repulsion and the alpha ray and nucleus so that path is deviated. It can also be concluded that the size of the nucleus is very small when compared with the size of the atom. That is why out of so many positive rays only very few are reflected.

As there is repulsion between the two positive charges, alpha particles cannot go and strike the nucleus. After they reach a certain distance, the repulsion dominates them so that either they deviates or reflects back. That’s why the particles are able to reach up to only a certain distance from the nucleus and this distance is called closest distance of approach. At that point all the kinetic energy of the alpha particle is converted into electrostatic potential energy between the Alpha particle and the protons of the nucleus.





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Problems and solutions on de Broglie's Hypothesis

Problem and solution

If 10,000 voltage is applied across an x-ray tube, what will be the ratio of the de  Broglie wavelengths of incident electron to that of the x-ray produced?

To solve this problem, we have to write the wavelength of the electron in terms of the de Broglie concept. Here in the place of the momentum we can write applied potential and the corresponding energy. Similarly to right the wavelength of the x-ray, we can use plank’s quantum concept as shown below.




Problems and solutions

Here we are going to solve two problems. In the first problem and alpha particle and the proton are passed through same magnetic field which is perpendicular to the velocity vectors. If both the charger particles are having the same radius, what is the ratio of their de Broglie wavelengths?

We can solve this problem by first of all understanding that the force due to the magnetic field on the charged particle provides the necessary centripetal force so that they takes a circular path. From that it can be concluded that the momentum of the particle is directly proportional to charge of the particle as well as the radius of the particle. Anyway in this problem being the radius is same, we can say momentum means directly proportional to the charge of the particle itself.

As per this concept, wavelength is inversely proportional to momentum that means wavelength is inversely proportional to the charge of the given particles.



The second problem is also solved basing on the same concept as shown above.

Problem and solution

Photons of energy 4.25 electron volt and 4.7 electron volt are allowed to incident on to metal surface S. If the maximum kinetic energies between them are having a difference of 1.5 electron volts and the they are wavelengths are in the ratio of 1:2, find the work functions of the two different metals?

We can express de Broglie wavelength in terms of the kinetic energy. Basing on this we can say that the ratio of the day Broglie wavelength is inversely proportional to Squire root of ratio of their respective kinetic energies. Taking this concept into consideration we can solve the problem as shown below.




Problem and solution

de Broglie wavelength of the proton accelerated  through a potential difference of 100 V is given. If a alpha particle is accelerated through same potential difference, what will be its wavelength?

According to de Broglie hypothesis, we know that the wavelength of a particle is the ratio of Planck’s constant to the momentum of the particle. We can express the momentum in terms of kinetic energy. Further we can express the kinetic energy in terms of the applied potential. By substituting the given data in that equation, we can solve the problem as shown below.




Problem and solution

A light particle of a certain mass at rest explodes into two particles having masses in the ratio of 2:3 . What is the corresponding ratio of the de Broglie wavelengths?

As the particle is in the state of rest, its initial momentum is equal to 0. The explosion is happened due to internal forces and hence law of conservation of momentum is very much valid here. According to this law the initial momentum of the system is equal to the final momentum when no external forces are acting on the system. As the momentum is as well as the Planck’s constant are same, the wavelengths are also going to be the same.



Problem and solution

What will be the wavelength of electron having energy of hundred electron volts?

We can solve this problem basing on dual nature of the particle. Here the particles shall be having a wave nature and hence it shall have certain wavelength which is equal to the ratio of Planck’s constant to momentum of the particle. We can express the momentum in terms of kinetic energy and the kinetic energy can be further expressed in terms of the applied voltage.

By substituting the data in the given problem we can solve the problem as shown below.



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