Rutherford’s Alpha() Particle Scattering
- The entire charge and almost entire mass of the atom is concentrated in a single core of the atom is called nucleus and is surrounded by Negatively(-) charged electron clouds.
- The particle and gold nuclei are so small that they may be treated as a point mass and point charge.
- The nucleus is considered so heavy that its motion during the scattering process may be neglected.
- The scattering is due to the coulombs force of repulsion between the particle and the Gold Nucleus.
- Each particle suffers a single deflection.
- The particles do not penetrate the nuclear region. So strong nuclear forces are not involved in their interaction.
Derivation Of Rutherford’s Scattering Formula
Let an particle is moving along PO apparatus and the relatively heavy nucleus stationary at it. Since both are Positively(+) charged so there is a force of repulsion between them. As particle gets closer to the nucleus the repulsive force highly increases and the particle follows a hyperbolic path. The asymptotes and po and p’o give the initial and final reflection of particle.
Let the perpendicular drawn from the nucleus to PO is NM and is the shortest distance between the nucleus and initial direction of 𝞪- particle which is known as impact parameters(b).
- Z be the atomic no. of the element That scattered particle.
- Ze be the charge of the nucleus.
- θ, be the scattering angle or angle of deviation of particle.
- As the particle moves its momentum changes from to i.e
using sine law in
Now From Equation (1)
where t = ,
t = ;
From Eq (2)
As the radius joining the — particle and the nucleus is along the direction of the force. So there is no torque acting on the — particle.
The angular momentum remain content
[ L=R *P =Rmv=bmv ]
putting this in EQ (3)
putting this EQ (4)
This is the relation between scatting angle and impact parameter.
In the actual experiment, a large number of particle are incident various impact parameter all around the (Au) Gold nucleus.
According to EQ(5), the particle that approaches the nucleus with impact parameter “b” will be scattered at an angle “ “
Particle approaching with the smallest value of “b” (i.e b) will be scattered through large angle (i.e )
The Area of cross-section of radius it.e is called scattering cross-section i.e
Consider a thin gold foil of thickness “t” and cross – section “A”. Suppose it contains ‘n’ number of atoms per unit volume, then the volume of the foil is “At”.
Then the number of target nuclei in the foil is “nAt”. We assume that the foil is so thin that particle suffers single deflection i.e , one particle is scattered by one gold nucleus.
Let ‘N’ be the total number of particle and “” be the Number of particle being scattered by an angle Then the fraction of incident particles scattered by an angle is
as we know from Eq(5)
But a particle detector measures the Number of df i.e,
Where N(0) represents the number of particle per unit area striking the screen. This is the expression for Rutherford’s Scattering formula.
Experimental Verification of Rutherford’s Alpha particle scattering Formula
The Theory of Rutherford’s Particle scattering was verified by a series of experiments conducted by Gaiger and Marsden in 1913.
The experimental arrangement consists of an Airtight chamber (c) which can be evacuated by tube (T). The chamber is capable of rotating inside the jacket ( J ) about a vertical axis. Radon is taken as the Radioactive source Rc of particles Which is placed inside a Radioactive cell (L).
After coming out of the narrow opening lead cavity the particle strikes a thin foil (F) (Gold, Silver, Platinum ).The foil is placed at the centre of the chamber (c).The scattered particles are viewed through low Power Microscope (M) which is provided with a fluorescent screen. As the chamber rotates about a central axis the microscope rotates along with it but the cavity (L) and foil(f) remain fixed with the tube. The particles striking on the foil get scattered in a different direction. The number of particles scattering along different direction can be recorded by observing the scintillations on the fluorescent screen.
The Following Results Obtained From The Above Experiment:-
- Most of the particle either passed straight through the metal foil or suffered only small deflection. This could be explained by Thomson’s Atomic Model.
- A few particle were deflected through angle where were less than 90 degree ( ) and a very few particle were deflected through an angle greater than 90 degree ( ). Sometimes a particle was found to be deflected through 180 degree( ) . These large angle scattering could not be explained by Thomson’s Atomic Model
- If is the scattering angle and N is the number of Particles available in that direction then it is found that
4.If “t” is the thickness of the foil and N is the number of particles scattered along scattering angle, It was observed that .This agreed with the theory.
5. This experiment also verified that Number of particles “N” scattered along scattering angle is directly proportional to () square of the Atomic Number of the foil atom i.e
Size of the nucleus from alpha() particles scattering:-
Initial Kinetic Energy of particle =
Initial Potential Energy of particle = 0
Final Kinetic Energy of particle = 0
Final Kinetic Energy of particle =
From energy conservation law,
Initial Total Energy = Final Total Energy
the charge of particle = 2e
the charge of nucleus = Ze
by putting the value of m,v and Z for given experiment we can find the radius of given nucleus.
Q. In an experiment velocity of particle is m/s is bombarded upon Gold Z= 79,
Mass of particle M= , Then what is the radius of the nucleus?
2. According to Rutherford’s mode electron can revolve in any orbit, So it must emit continuous radiation of all frequencies but elements emit spectral lines of only definite frequency.