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Alpha particles hit a gold nucleus head-on, at the same energy each time…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Alpha particles hit a gold nucleus head-on, at the same energy each time. (a) The energy of the alpha particles is doubled. By what factor does the closest approach change? (b) The target is changed from gold, $Z = 79$ , to aluminium, $Z = 13$ , at the first energy. By what factor does the closest approach change?

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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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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