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An alpha particle is fired straight at a lead nucleus, with Z = 82 and A = 208…

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An alpha particle is fired straight at a lead nucleus, with $Z = 82$ and $A = 208$ . What energy must it have to reach the surface of the nucleus, which has a radius of $1.2\,A^{1/3}$ fm? Find the radius first. Take $\dfrac{e^2}{4\pi\varepsilon_0} = 1.44\ \text{MeV fm}$ .

Need a reminder? The key ideaShow

In a head-on hit the alpha particle stops where its kinetic energy has become electric potential energy. With $K$ in MeV and $\dfrac{e^2}{4\pi\varepsilon_0} = 1.44\ \text{MeV fm}$ :

\displaystyle r_0 = \frac{2Ze^2}{4\pi\varepsilon_0 K} = \frac{2Z \times 1.44}{K}\ \text{fm}

A nucleus with mass number $A$ has a radius of about $1.2\,A^{1/3}$ fm.

The worked solution

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