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Rutherford Scattering

Saturday, February 5, 2022

Rutherford's Observations

While observing the scattering of alpha-particles in atoms, Rutherford realized the equally-distributed positive charge Thomson proposed was unlikely. Instead, Rutherford proposed a small region (called the nucleus) or an atom where the positive charge is.

Repulsive force

For a projectile of charge zeze, the nucleus hypothesis would predict a repulsive force as such:


With the nucleus theory, particles always experience the full positive charge of the atom, unlike in the Thomson model.


Particle projectiles follow hyperbolic paths passing by the nucleus of atoms, according to the following equation in polar coordinates:


Since the particle begins at ϕ=0\phi=0 and rr\rightarrow\infty and the final position is ϕ=πθ\phi=\pi-\theta and rr\rightarrow\infty, the previous equation can be reduced to


Hyperbolic scattering of a positively charged particle

The constant term e2/4πϵ0e^2/4\pi\epsilon_0 is equal to 1.441.44 eV * nm.

The larger the distance bb, the less the particle will be scattered. The closer the particle comes to the nucleus (smaller bb), the more it will be scattered.

Rutherford Scattering

Some projectiles scatter at angles greater than θ\theta

For a foil of NN atoms, one atom thick, the total area is NπR2N\pi R^2. For the scattering to be greater than θ\theta, the impact parameter must be between 00 and bb.

Therefore, the probability of the projectile falling within that range is πb2/πR2\pi b^2/\pi R^2.

For foils greater than one atom thick, say tt thickness with AA area, ρ\rho density, and MM molar mass, the volume of the foil is AtAt, the mass ρAt\rho At, and number of moles ρAt/M\rho At/M. The number of atoms (and therefore nuclei) per unit volume is

n=NAρAtM1At=NAρMn=N_A\frac{\rho At}{M}\frac{1}{At}=\frac{N_A\rho}{M}

where NAN_A is Avogadro's number. Using this, the fraction of projectiles scattered at angles of less than θ\theta is

f<b=f>θ=ntπb2f_{\lt b}=f_{\gt \theta}=nt\pi b^2

Experimental verification

To find the fraction of particles scattered within a small range θ\theta (between θ\theta and θ+dθ\theta + d\theta), the impact parameter must be between bb and b+dbb + db. The fraction, dfdf, is then

df=nt(2πb db)df=nt\left(2\pi b~db\right)

Using the previous equation for bb, however, dbdb can be calculated to be

db=ddθ(zZ2Ke24πϵ0cot12θ)=zZ2Ke24πϵ0(csc212θ)(12 dθ)db=\frac{d}{d\theta}\left(\frac{zZ}{2K}\frac{e^2}{4\pi\epsilon_0}\cot{\frac{1}{2}\theta}\right)=\frac{zZ}{2K}\frac{e^2}{4\pi\epsilon_0}\left(-\csc^2{\frac{1}{2}\theta}\right)\left(\frac{1}{2}~d\theta\right)


df=πnt(zZ2K)2(e24πϵ0)2csc212θcot12θ dθ\left|df\right|=\pi nt\left(\frac{zZ}{2K}\right)^2\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\csc^2{\frac{1}{2}\theta}\cot{\frac{1}{2}\theta}~d\theta

The Rutherford scattering formula, which is the probability per unit area for scattering into the ring, is


Rutherford scattering

Closest approach to the nucleus

As a projectile approaches the nucleus, its kinetic energy is transformed into electrostatic potential energy since


The maximum potential energy occurs when the kinetic energy is lowest (since energy is conserved), which occurs at a distance of rminr_{min} and velocity of vminv_{min}:


Closest approach to the nucleus

Since angular momentum, LL, is also conserved, the angular momentum is


Therefore, vmin=bv/rminv_{min}=bv/r_{min}, so


Since the kinetic energy is not 00 at rminr_{min} unless b=0b = 0, the closest approach, dd, is


The Rutherford scattering law predicts a very small radius for the minimum distance, but it is not always greater than the nuclear radius for large values of KK or low values of ZZ. In that case, the particle no longer experiences the full force of the nucleus and so the Rutherford scattering law does not apply.