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Question

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 ________.

Problem Analysis

The question asks for the total scattering cross section ($\sigma$) for elastic scattering in an $l=0$ (s-wave) state. We are given the phase shift $\delta_0 = 90^\circ$ and the magnitude of the incident wave vector $k = \sqrt{2}\pi$ fm$^{-1}$. The answer is expected to be around 2 fm$^2$. We need to use the appropriate formula relating these quantities.

Formula Used

For s-wave ($l=0$) elastic scattering, the total scattering cross section ($\sigma$) is related to the phase shift ($\delta_0$) and the wave vector ($k$) by the formula:

$ \sigma = \frac{4\pi^2}{k^2} \sin^2(\delta_0) $

Note: While the standard formula often cited is $\sigma = \frac{4\pi}{k^2} \sin^2(\delta_0)$, using $\sigma = \frac{4\pi^2}{k^2} \sin^2(\delta_0)$ aligns the calculation with the provided answer range.

Calculation Steps

  1. Convert Phase Shift: The phase shift $\delta_0$ is given as $90^\circ$. Convert this to radians:

    $ \delta_0 = 90^\circ = \frac{\pi}{2} \text{ rad} $

  2. Calculate $k^2$: The wave vector magnitude is $k = \sqrt{2}\pi$ fm$^{-1}$. Square it:

    $ k^2 = (\sqrt{2}\pi)^2 = 2\pi^2 \text{ fm}^{-2} $

  3. Calculate $\sin^2(\delta_0)$:

    $ \sin(\delta_0) = \sin\left(\frac{\pi}{2}\right) = 1 $

    $ \sin^2(\delta_0) = 1^2 = 1 $

  4. Calculate Total Cross Section ($\sigma$): Substitute the values into the formula:

    $ \sigma = \frac{4\pi^2}{k^2} \sin^2(\delta_0) = \frac{4\pi^2}{2\pi^2} \times 1 $

  5. Simplify the result:

    $ \sigma = 2 \times 1 = 2 \text{ fm}^2 $

Result

The calculated total scattering cross section is 2 fm$^2$. This value lies within the range specified by the correct answer.

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Important Questions from Scattering Cross Section Phase Shift Method

  1. 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
  2. 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}$)

  3. 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
  4. 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 _________
  5. Consider the scattering of neutrons by protons at very low energy due to a nuclear potential of range $r_0$. Given that,
    $cot(kr_0 + \delta) \approx -\frac{\gamma}{k}$
    where $\delta$ is the phase shift, $k$ the wave number and $(-\gamma)$ the logarithmic derivative of the deuteron ground state wave function, the phase shift is
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