$\left(k = \frac{1}{4\pi\epsilon_0} \text{ and } r_0 = k \frac{Z e^2}{m v_0^2}\right)$
Alpha Particle Scattering: Angular Momentum Conservation
The interaction between the $\alpha$ particle and the nucleus is governed by the Coulomb force. This force is a central force, meaning it always acts along the line connecting the two particles. A key consequence of a central force is the conservation of angular momentum.
Derivation of Angular Momentum Conservation
Let $m$ be the mass of the $\alpha$ particle.
- Initial State (Far from Nucleus):
The $\alpha$ particle approaches the nucleus with an initial speed $v_0$ and an impact parameter $b$. The angular momentum ($L_i$) relative to the nucleus is given by:
$ L_i = m v_0 b $
- State at Closest Approach ($r_m$):
At the distance of closest approach $r_m$, the velocity $v_m$ of the $\alpha$ particle is momentarily perpendicular to the position vector connecting it to the nucleus. The angular momentum ($L_f$) at this point is:
$ L_f = m v_m r_m $
Applying Conservation Law
Since angular momentum is conserved during the scattering process:
$ L_i = L_f $
Substituting the expressions for $L_i$ and $L_f$:
$ m v_0 b = m v_m r_m $
Dividing both sides by the mass $m$ gives:
$ v_0 b = v_m r_m $
This relationship holds true based on the conservation of angular momentum. Therefore, the option stating $v_0 b = v_m r_m$ is correct.
The other options involve considerations of energy conservation and approximations under specific conditions, such as when the impact parameter is much smaller than the characteristic length scale $r_0$.
Important Questions from Scattering Cross Section Phase Shift Method
- A particle is scattered from a potential $ V(\vec{r}) = g\delta^3(\vec{r}) $, where $ g $ is a positive constant. Using the first Born approximation, the angular $ (\theta, \phi) $ dependence of differential scattering cross section $ \frac{d\sigma}{d\Omega} $ is
Consider the potential $U(r)$ defined as $$U(r) = -U_0 \frac{e^{- \alpha r}}{r}$$ where $ \alpha$ and $U_0$ are real constants of appropriate dimensions. According to the first Born approximation, the elastic scattering amplitude calculated with $U(r)$ for a (wave-vector) momentum transfer $q$ and $ \alpha \to 0$, is proportional to
(Useful integral: $ \int_0^{ \infty} \sin(qr)e^{- \alpha r} dr = \frac{q}{ \alpha^2+q^2}$)
- The scattering of particles by a potential can be analyzed by Born approximation. In particular, if the scattered wave is replaced by an appropriate plane wave, the corresponding Born approximation is known as the first Born approximation. Such an approximation is valid for
- Consider an elastic scattering of particles in $l = 0$ states. If the corresponding phase shift $\delta_0$ is $90^\circ$ and the magnitude of the incident wave vector is equal to $\sqrt{2}\pi$ fm$^{-1}$ then the total scattering cross section in units of fm$^2$ is ________.
- Protons and $\alpha$-particles of equal initial momenta are scattered off a gold foil in a Rutherford scattering experiment. The scattering cross sections for proton on gold and $\alpha$-particle on gold are $\sigma_p$ and $\sigma_\alpha$ respectively. The ratio $\sigma_\alpha/\sigma_p$ is _________
