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Alpha particles of energy 7 MeV are incident from a gold foil in a scattering experiment.
- Plot the electrostatic potential seen by the alpha particles
- Find the closest distance an alpha particle can reach to the nucleus assuming that initially the alpha particle was on a direct collision course with the gold nucleus.
How will your answers change if the positive charge of the gold nucleus is assumed to be uniformly distributed over a sphere of radius of a few Angstroms, as was proposed in Thompson model?
Answer:- \(r_\text{min} \approx 32 \text{fm}\) ...............................................................................
Solution:
\begin{eqnarray} r &=& \frac{2Ze^2}{4\pi\epsilon_0 E} \nonumber\\ &=& \Big(\frac{2Ze^2}{4\pi\epsilon_0\hbar {c}}\Big) \Big(\frac{m_\pi c^2}{E}\Big)\Big(\frac{\hbar}{m_\pi c}\Big) \nonumber\\ &=&\Big( \frac{2\times79}{137}\Big)\Big(\frac{140}{7}\Big) \times 1.4 \text{ fm} \nonumber\\ &\approx&\frac{2\times 80 \times 1.4}{7} = 32 \text{ fm} \end{eqnarray}
NOTE : We have used
- Pion pass \(\m_pi c^2=140\) MeV,
- Compton wavelength of pion =\(\hbar/(m_\pi c^2) =1.4 fm\)
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4727:Diamond Point
