January 2018 ~ The Science Thinkers

Rutherford's Alpha( $\alpha-$ ) Particle Scattering

The theory of rutherford's $\alpha-$ particle scattering is based on the

Following Assumption:-

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 $\alpha-$ 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 $\alpha-$ particle and the Gold Nucleus.
Each $\alpha-$ particle suffers a single deflection.
The $\alpha-$ 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 $\alpha-$ 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 $\alpha-$ particle gets closer to the nucleus the repulsive force highly increases and the $\alpha-$ particle follows a hyperbolic path. The asymptotes and po and p'o give the initial and final reflection of $\alpha-$ 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).

Let

Z be the atomic no. of the element That scattered $\alpha-$ particle.
Ze be the charge of the nucleus.
θ, be the scattering angle or angle of deviation of $\alpha-$ particle.
As the $\alpha-$ particle moves its momentum changes from $\vec{P_{1}}$ to $\vec{P_{2}}$ i.e $\vec{\Delta P}= \vec{P_{2}}-\vec{P_{1}}=\int \vec{F}dt------(1)$

using sine law in $\Delta zxy$

$\frac{yz}{sin\Theta }=\frac{xz}{sin\frac{\pi -\theta }{2}}$
$\Rightarrow \frac{\Delta P}{sin\theta }=\frac{\vec{P_{2}}}{sin(\frac{\pi }{2}-\frac{\theta }{2})}$

$\Rightarrow \frac{\Delta P}{sin \theta }=\frac{mv}{cos(\frac{\theta}{2})}$

$\Rightarrow \Delta P=\frac{mvsin\theta}{cos(\frac{\theta}{2})}$

$\Rightarrow \Delta P=\frac{2mv . Sin(\frac{\theta}{2}) cos(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

${\color{Red}\Rightarrow \Delta P= 2.mv.sin(\frac{\theta}{2})........(2) }$

Now From Equation (1)

$\Rightarrow \Delta P =\int_{-\infty }^{\infty}Fdt$

$\Delta P=\int_{-\infty }^{\infty }F.cos\phi.dt$

where t = $- \infty$ ,    $\phi= -(\frac{\pi -\theta }{2})$

t = $\infty$ ; $\phi = (\frac{\pi -\theta }{2})$
$\Rightarrow \Delta P = \int_{-(\frac{\pi -\theta }{2})}^{\frac{\pi -\theta }{2}} Fcos\phi dt$

$\Rightarrow \Delta P = \int_{-(\frac{\pi -\theta }{2})}^{\frac{\pi -\theta }{2}} Fcos\phi \frac{dt}{d\phi }.d\phi$

From Eq (2)

${\color{Red} \Rightarrow 2mvsin\frac{\theta }{2} = \int_{-(\frac{\pi -\theta }{2})}^{\frac{\pi -\theta }{2}} Fcos\phi \frac{dt}{d\phi }.d\phi--------------(3)}$

As the radius   $\vec{b}$ joining the $\alpha$ -- particle and the nucleus is along the direction of the force. So there is no torque acting on the $\alpha$ -- particle.

The angular momentum remain content

$\Rightarrow I\omega =L$

$\Rightarrow m r^{2}\omega =bmv$ [ L=R *P =Rmv=bmv ]

$\Rightarrow \frac{d\theta }{dt}=\frac{bv}{r^{2}}$

$\Rightarrow \frac{dt}{d\theta}=\frac{r^{2}}{bv}$

putting this   $\theta$ in EQ (3)

${\color{Red} 2mv sin\frac{\theta }{2}=\int_{-(\frac{\pi -\theta }{2})}^{\frac{\pi -\theta }{2}}Fr^{2}cos\phi d\phi ---------(4)}$

But    $F=\frac{1}{r\pi \epsilon_{0} }\frac{Ze2e}{r^{2}}$

   $\Rightarrow F=\frac{1}{4\pi\epsilon _{0}}\frac{2Ze^{2}}{r^{2}}$

putting this EQ (4)

$\Rightarrow 2mbv^{2}sin(\frac{\theta }{2})= \int_{-(\frac{\pi -\theta }{2})}^{\frac{\pi -\theta }{2}}\frac{1}{4\pi \epsilon _{0}}\frac{2Ze^{2}}{r^{2}} r^{2}cos\phi d\phi$

$\Rightarrow mbv^{2}\sin \frac{\theta }{2}=\frac{Ze^{2}}{4\pi \epsilon _{0}}\int_{-(\frac{\pi -\theta }{2})}^{(\frac{\pi -\theta }{2})}\cos \phi d\phi$

$\Rightarrow mbv^{2}\sin (\frac{\theta }{2})=\frac{Ze^{2}}{4\pi \epsilon _{0}}\left [ \sin (\frac{\pi -\theta }{2})-\sin- (\frac{\pi -\theta }{2}) \right ]$

$\Rightarrow mbv^{2}\sin (\frac{\theta }{2})=\frac{Ze^{2}}{4\pi \epsilon _{0}}\left ( \cos \frac{\theta }{2} +\cos \frac{\theta }{2}\right )$

$\Rightarrow mbv^{2}\sin (\frac{\theta }{2})=\frac{Ze^{2}}{4\pi \epsilon _{0}}2\cos \frac{\theta }{2}$

$\Rightarrow \frac{mbv^{2}.4\pi \epsilon _{0}}{2Ze^{2}}=\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}=\cot \frac{\theta }{2}$

$\Rightarrow \cot \frac{\theta }{2}=\frac{2\pi \epsilon _{0}mv^{2}b}{Ze^{2}}=\frac{2\pi \epsilon _{0}.2.\frac{1}{2}.mv^{2}b}{Ze^{2}}$

${\color{DarkRed} \Rightarrow \cot \frac{\theta }{2}=\frac{4\pi \epsilon _{0}kb}{Ze^{2}}--------------(5)}$    where k=K.E

This is the relation between scatting angle and impact parameter.

In the actual experiment, a large number of $\alpha-$ particle are incident various impact parameter all around the (Au) Gold nucleus.
According to EQ(5), the $\alpha-$ particle that approaches the nucleus with impact parameter "b" will be scattered at an angle " $\theta$ "

$\alpha-$ Particle approaching with the smallest value of "b" (i.e $\leq$ b) will be scattered through large angle (i.e $\geq \theta$ )
The Area of cross-section of radius it.e $\pi b^{2}$ is called scattering cross-section i.e   ${\color{Red} \sigma = \pi b^{2}}$
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 $\alpha-$ particle suffers single deflection i.e , one $\alpha-$ particle is scattered by one gold nucleus.

Since one nucleus has cross - section $\LARGE \sigma$ .
So area of cross - section of "nAt" nuclei = $\LARGE \sigma nAt$ .

Let 'N' be the total number of $\alpha-$ particle and " $\large N_{0}$ " be the Number of $\alpha-$ particle being scattered by an angle $\geq \theta$ Then the fraction of incident $\alpha-$ particles scattered by an angle $\geq \theta$ is

$\large f=\frac{N}{N_{0}}=\frac{target\: area\: exposed\: by\: scattere}{total\: target\: area}$

$\large \Rightarrow f=\frac{\sigma nAt}{A}$

$\large \Rightarrow f=\sigma nt$

$\large \Rightarrow f=\pi b^{2}nt$ as $\large \sigma =\pi b^{2}$

$\large \Rightarrow f=\pi nta^{2}cot^{2}\frac{\theta }{2}$
-------------------------------------------
as we know from Eq(5)
   $\large \left | \cot\frac{\theta }{2}=\frac{4\pi \epsilon _{0}kb} {Ze^{2}}\Rightarrow b=\frac{Ze^{2}cot\frac{\theta }{2}}{4\pi \epsilon _{0}k} \right |$

   $\large \left | \Rightarrow b=a\cot\frac{\theta }{2} \left ( a=\frac{Ze^{2}}{4\pi \epsilon _{0}k} \right ) \right |$
-------------------------------------------
But a particle detector measures the Number of df i.e,

$df=\frac{dN}{N_{0}}$

$\Rightarrow df=\pi nta^{2}.2cot\frac{\theta }{2}(-cosec^{2}\frac{\theta }{2}).\frac{1}{2}d\theta$

$\Rightarrow df=-\pi nta^{2} \frac{cot\frac{\theta }{2}}{sin^{2}\frac{\theta }{2}} d\theta$

${\color{Red} \Rightarrow \frac{dN}{N_{0}}=-\pi nta^{2} \frac{cot\frac{\theta }{2}}{sin^{2}\frac{\theta }{2}} d\theta }$

The scattered   $\alpha-$ particle strikes the screen placed at a distance "r" from the foil. The area ds is struck by these particles is the area of the ring of radius rsin $\theta$ and thickness " $rd\theta$ " is
$ds= 2\pi (rsin\theta )rd\theta =2\pi r^{2}sin\theta d\theta$

$ds=2\pi r^{2}2sin\frac{\theta }{2}cos\frac{\theta }{2} d\theta$

$\Rightarrow ds=4\pi r^{2}sin\frac{\theta }{2}cos\frac{\theta }{2} d\theta$

Then    $\frac{dF}{ds}=\frac{dN}{N_{0}ds}$

$\Rightarrow \frac{dF}{ds}=\frac{-\pi nta^{2}\frac{cot\frac{\theta }{2}}{sin^{2}\frac{\theta}{2}}d\theta}{N_{0}4\pi r^{2}sin\frac{\theta}{2}.cos\frac{\theta}{2}}$ $\left | \frac {cot\frac{\theta}{2}}{sin^{2}\frac{\theta}{2}}=\frac{\frac{cos\frac{\theta}{2}}{sin\frac{\theta}{2}}}{sin^2\frac{\theta}{2}}=\frac{cos\frac{\theta}{2}}{sin^{3}\frac{\theta}{2}} \right |$

$\Rightarrow \frac{dN}{N_{0}ds}= \frac{-nta^{2}cos\frac{\theta}{2}}{4r^{2}sin^{4}\frac{\theta}{2}.cos\frac{\theta}{2}}$

$\Rightarrow \frac{dN}{N_{0}ds}= \frac{-nta^{2}}{4r^{2}sin^{4}\frac{\theta}{2}}$

$\Rightarrow \frac{dN}{N_{0}ds}= \frac{-nt}{4r^{2}sin^{4}\frac{\theta}{2}}\times \frac{Z^{2}e^{4}}{16\pi ^{2}\varepsilon _{0}^{2}k^{2}}$ $\left ( \therefore a=\frac{Ze^{2}}{4\pi \epsilon _{0}k} \right )$

$\Rightarrow \frac{dN}{ds}=\frac{-N_{0}nt.\: Z^{2}e^{4}}{(8\pi \epsilon _{0})^{2}r^{2}k^{2}sin^{4}\frac{\theta}{2}}$

$\large \Rightarrow\left | \frac{dN}{ds}\right |=N(0)=\frac{N_{0}nt.\: Z^{2}e^{4}}{(8\pi \epsilon _{0})^{2}r^{2}k^{2}sin^{4}\frac{\theta}{2}}$

Where N(0) represents the number of   $\alpha-$ 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 $\alpha-$ 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 $\alpha-$ particles Which is placed inside a Radioactive cell (L).

After coming out of the narrow opening lead cavity the $\alpha-$ particle strikes a thin foil (F) (Gold, Silver, Platinum ).The foil is placed at the centre of the chamber (c).The scattered $\alpha-$ 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 $\alpha-$ particles striking on the foil get scattered in a different direction. The number of $\alpha-$ 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 $\alpha-$ particle either passed straight through the metal foil or suffered only small deflection. This could be explained by Thomson's Atomic Model.
A few $\alpha-$ particle were deflected through angle where were less than 90 degree ( $\leqslant 90^{0}$ ) and a very few $\alpha-$ particle were deflected through an angle greater than 90 degree ( $> 90^{0}$ ). Sometimes a particle was found to be deflected through 180 degree( $180^{0}$ ) . These large angle scattering could not be explained by Thomson's Atomic Model
If $\large \theta$ is the scattering angle and N is the number of Particles available in that direction then it is found that

$\large \log_{e}N\propto \frac{1}{\sin ^{4}\frac{\theta }{2}}\propto cosec^{4}\frac{\theta }{2}$

which is perfectly agree with the theory.
4.If "t" is the thickness of the foil and N is the number of $\alpha-$ particles scattered along scattering angle $\large \theta$ , It was observed that $N\propto t$ .This agreed with the theory.

5. This experiment also verified that Number of $\alpha-$ particles "N" scattered along scattering angle $\large \theta$ is directly proportional to ( $\propto$ ) square of the Atomic Number of the foil atom i.e $N\propto Z^{2}$

Size of the nucleus from alpha( $\alpha-$ ) particles scattering:-

Let an $\alpha-$ particle having velocity $\vartheta$ approach a nucleus (head on) having a charge (+Ze). The velocity of   $\alpha-$ particle decreases till it comes to rest at a distance "b" from the nucleus.It is the then repelled back along the direction of approach.

Initial Kinetic Energy of   $\alpha-$ particle = $\frac{1}{2}mv^{2}$

Initial Potential Energy of   $\alpha-$ particle = 0

Final Kinetic Energy of   $\alpha-$ particle = 0

Final Kinetic Energy of $\alpha-$ particle =   $\large \frac{1}{4\pi \epsilon _{0}}\, \frac{q_{1}q_{2}}{r_{0}}$

From energy conservation law,

Initial Total Energy = Final Total Energy

$\large \Rightarrow \frac{1}{2}mv^{2}=\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{r_{0}}$

$\large \Rightarrow \frac{1}{2}mv^{2}=\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{r_{0}}$

$\large \Rightarrow r_{0}= \frac{1}{4\pi \epsilon _{0}}\frac{2q_{1}q_{2}}{mv^{2}}$

the charge of $\alpha-$ particle = 2e
the charge of nucleus = Ze

$\large \Rightarrow r_{0}= \frac{2}{4\pi \epsilon _{0}}.\frac{2e\times Ze}{mv^{2}}= \frac{1}{4\pi \epsilon _{0}}\frac{4Ze^{2}}{mv^{2}}$
$\large \Rightarrow r_{0}= (9\times 10^{9}).\frac{4Z(1.6\times 10^{-19})^{2}}{mv^{2}}$
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   $\alpha-$ particle is   $\large 2\times 10^{7}$ m/s is bombarded upon Gold Z= 79,
Mass of   $\alpha-$ particle M= $\large 4\times 1.67\times 10^{-27} g$ , Then what is the radius of the nucleus?

$\large r_{0}= \frac{4(9\times 10^{9})\times 79\times (1.6\times 10^{-19})^{2}}{4\times (1.67)\times 10^{-27}\times (2\times 10^{7})^{2}}=2.62\times 10^{-14}m$

Failure of Rutherford's scattering formula

1. According to electromagnetic theory, a charged particle in accelerated motion must radiate energy in the form of electromagnetic radiation. As a result, there should be a gradual decrease in the energy of the electron. The electron should follow a spiral path and ultimately fall into the nucleus. thus the whole atomic structure should collapse. this is contrary to the actual fact that atom is very stable.

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.

The Science Thinkers

Rutherford's Alpha particle scattering Experiment

Rutherford's Alpha( $\alpha-$ ) Particle Scattering

Derivation Of Rutherford's Scattering Formula

Experimental Verification of Rutherford's Alpha particle scattering Formula

The Following Results Obtained From The Above Experiment:-

Size of the nucleus from alpha( $\alpha-$ ) particles scattering:-

Failure of Rutherford's scattering formula

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Rutherford's Alpha particle scattering Experiment

Rutherford's Alpha() Particle Scattering

Derivation Of Rutherford's Scattering Formula

Experimental Verification of Rutherford's Alpha particle scattering Formula

The Following Results Obtained From The Above Experiment:-

Size of the nucleus from alpha() particles scattering:-

Failure of Rutherford's scattering formula

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EVEN STAR SYSTEMS HAVE IDENTITY CRISES

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Rutherford's Alpha( $\alpha-$ ) Particle Scattering

Size of the nucleus from alpha( $\alpha-$ ) particles scattering:-