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
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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} $
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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} $
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Calculate $\sin^2(\delta_0)$:
$ \sin(\delta_0) = \sin\left(\frac{\pi}{2}\right) = 1 $
$ \sin^2(\delta_0) = 1^2 = 1 $
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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 $
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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.
Important Questions from Scattering Cross Section Phase Shift Method
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(Useful integral: $ \int_0^{ \infty} \sin(qr)e^{- \alpha r} dr = \frac{q}{ \alpha^2+q^2}$)
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