# Alpha Particle Scattering and the Nuclear Model of the Rutherford Atom - GeeksforGeeks (2023)

In today's universe, the smallest particle of matter is an atom. It is the smallest possible result obtained by the division of matter without releasing electrically charged particles. It was first proposed by John Dalton on behalf of atomic theory. This theory has its own development suggested by several scientists such as J.J. Thompson, Ernest Rutherford, Niels Bohr and Erwin Schrödinger. Today, Rutherford's model of the atom will be discussed.

### Rutherford Model

Ernest Rutherford worked on the emission of α particles by atoms to understand the structure of atoms, and regarding the experiment, Rutherford Geiger and Marsden did some experiments of their own, as shown in the figure. Experimental setup for the Geiger-Marsden experiment

They used a 5.5 MeV beam emitted by a radioactive source (bismuth) on a thin sheet of gold. The α particles emitted by the source are concentrated into a single beam of light incident on the object's gold foil, and the concentrated beam is scattered upon impact and can be seen on the screen through a microscope.

This experiment was carried out and the result was that in the scattered beam of light, many particles pass through the sheet without collusion, only 0.14% of the incident particles are scattered more than 1oand 1 in 8,000 particles deflect more than 90o. This means that the particle has completely dispersed in the reverse direction, which means that there must be some kind of large repulsive force. Why did this distract 1 in 8000?

Rutherford said from the observation that the greatest repulsive force can be provided when mass and positive charge are concentrated at the center, causing reflection.

Thus, in Rutherford's atomic model, all the positive charge and mass are at the center, and the electrons move around it like planets at some distance from the sun.

Rutherford said the size of the atom will be 10-15more or less, but actually there were 10-10and when the electrons are some distance away, the atom is mostly empty space. Due to a large empty space in an atom, it is obvious that most α particles pass directly through the atom, and if any α particles approach the nucleus, they will be scattered due to the electric field.

### α particle dispersion

As we all know, gold foil is very thin, so we can assume that α particles will not suffer more than one collision. Therefore, the calculation was necessary for a single core. The α particles are nuclei of a helium atom and therefore carry 2 units (2e). The charge in the gold nucleus is Ze, where Z is the atomic number of gold.

In this experiment, the gold particles are much heavier than an α particle, the gold particles do not move when scattering occurs. The trajectory can be calculated using Newton's second law of motion and Coulomb's law of electrostatic repulsion between α particles and positively charged nuclei, which can be written as:

F = 1/(4p0) × (2e)(Ze)/r2

Oh,

• r = distance between as particles α e o nucleus.
• Ze = gold core charge

### Alpha particle trajectory The trajectory of the alpha particle depends on the impact parameter (the perpendicular distance of the initial velocity vector of the α particle from the center of the nucleus. In the figure above there are different probabilities for the dispersion of the alpha particle and how it can look at this is close to the α particle of the nucleus, the one that suffers the greatest scattering, since the α particle that hits the nucleus directly can suffer a scattering of up to 180 degrees.

But, by observation, it is evident that few particles experience this 180 degree deflection because the number of particles returned was significant.

### Electron orbits in the Rutherford nuclear model

The model incorporates classic concepts that

• It appears that the atom is an electrically neutral sphere.
• It has a very small, massive, positively charged nucleus at its center.
• The nucleus in the center is surrounded by orbiting electrons.
• These electrons revolve around the nucleus in their respective orbits.
• The electrostatic attraction (Fmi) between the orbiting electrons and the nucleus provides the necessary centripetal force (FC) to keep the electrons spinning around the nucleus]

So this can be represented as,

Fmi-FC

[1/(4po)] × [z2/R2] = mv2/R

such is the relationship between the radius of the orbit and the speed of the electron

r = mi2/(4pom.v.2)

As we all know, kinetic energy (k) = 1/2 × (mv2)

then the kinetic energy,

k = z2/(8poR)

Likewise, the potential energy (U) increases

U = -e2/(4po)

Therefore, the total energy becomes the sum of the kinetic and potential energies.

T = K + U = [z2/(8por) + (-z2)/(4poR))]

T = -e2/(8poR)

Therefore, the total energy of the electrons is negative, which means that the electrons are electrically bound to the nucleus.

### examples of problems

Problem 1: In Rutherford's nuclear model of the atom, the nucleus is analogous to the sun around which electrons orbit (radius ≅ 10-10m) how the earth revolves around the sun. If the dimensions of the solar system were the same as the proportions of the atom, would the Earth be closer or farther from the Sun than it actually is? The radius of Earth's orbit is about 1.5 × 1011M. The radius of the sun is supposed to be 8 × 108

Solution:

The ratio of the orbital radius of the electron to the radius of the nucleus is (10-10m/10-15m) = 105

Therefore, it can be seen that the radius of the orbit of the electron is 105times greater than the radius of the nucleus.

If we consider the same situation for the Earth and the Sun, that is, the Earth's orbit around the Sun was 105times greater than the radius of the sun, then the radius of the earth will comply with this condition

105× 8 × 108m = 8 × 1013m, which is certainly longer than Earth's actual orbit, and therefore Earth would be much farther from the Sun.

It also implies that an atom contains a much larger proportion of empty space compared to our solar system.

Problem 2: It has been discovered experimentally that 13.6 eV of energy is required to split a hydrogen atom into a proton and an electron. Calculate the orbital radius and speed of the electron in a hydrogen atom.

Solution:

The total energy of the electron in the hydrogen atom is -13.6 eV = -13.6 × 1.6 × 10-19j

According to the relationship between the radius of the orbit and the velocity of the electron we obtain,

mi = -e2/(8poR)

E = -2,2 × 10-18j

Therefore, the radius of the orbit increases

r = -e2/(8poMI)

r = [(9 × 109New Mexico2/C2) (1,6 × 10-19C)2] / [(2) (-2,2 × 10-18j)]

r = 5,3 × 10-11METRO.

This allows us to calculate the velocity of the electron.

v = mi / (√4 poSir)

v = 2,2 × 106EM

### conceptual issues

Question 1: What radioactive source was used in the Geiger and Marsden alpha particle scattering experiment?

Responder:

The radioactive element used in the alpha scattering experiment is a radioactive isotope ofbismuth(214 Bi 83).

Question 2: What was the deviation ratio in the alpha scatter experiment for 90o?

Responder:

1 in 8000 of alpha particles deviated more than 90o.

Question 3: Explain the behavior of alpha particles when they hit the gold leaf in the alpha scattering experiment.

Responder:

When the alpha particles from the source hit the gold foil, many of them pass directly through the gold foil, only about 0.14% of the particles that hit the core are scattered more than 1o, 1 in 8,000 particles deflect more than 90° and very few hit the nucleus head-on and return towards the source, i.e. h 180o.

Question 4: What are the two laws used to calculate the trajectory of the alpha particle that collided?

Responder:

Newton's second law of motion and Coulomb's law of electrostatic repulsion between α particles and the positively charged nucleus.

Question 5: Give any two properties of an atom given by the Rutherford nuclear model.

Responder:

The properties of an atom given by the Rutherford nuclear model are:

1. The nucleus in the center is surrounded by orbiting electrons.
2. These electrons revolve around the nucleus in their respective orbits.

Question 6: Why did alpha particles change their trajectory when they collided with the nucleus? Why not alpha particles?

Responder:

Alpha particles are much lighter than gold particles. If two substances have different masses and the lighter one collides with the heavier one, the lighter one will be knocked over.

Question 7: Why is the trajectory different for particles colliding in different places in the nucleus?

Responder:

According to the observation of the scattering experiment, the nucleus has a repulsive force and is concentrated in the center. Thus, when the electron collides with the head, it is deflected backwards 180oand when the particles collide further above or below the center, or even closer to the nucleus, they are deflected by their respective angles.

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