Question

In a head-on collision, the closest approach of a 6.24 MeV alpha particle to the center of a nucleus is 6.00 fm. The nucleus is in an atom of what element? Assume the nucleus remains at rest.

Short answer

The element is Aluminum

Step-by-step solution

Step 1. Given information is :Kinetic energy of alpha particle = 6.24 MeVClosest approach = 6.00 fm

We need to find out the element used.

Step 2. Comparing Kinetic and potential energy for the closest approach

Let Z be the atomic number of the element.

$$\begin{gathered} Chargeonelement=Z\times 1.6\times {10}^{-19}C \\ Chargeonalphaparticle=2\times 1.6\times {10}^{-19}C \\ Dis\tan ceofclosestapproach=6.00fm=6.00\times {10}^{-15}m \\ Kineticenergyofalphaparticle=6.24MeV \\ =6.24\times {10}^6\times 1.6\times {10}^{-19}C \end{gathered}$$

At the closest approach, Kinetic energy will be equal to the potential energy. Therefore,

$$\begin{gathered} \frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}\frac{q_1{q}_2}{r}=K \\ 9\times {10}^9\times \frac{Z\times 1.6\times {10}^{-19}\times 2\times 1.6\times {10}^{-19}}{6.0\times {10}^{-15}}=6.24\times {10}^6\times 1.6\times {10}^{-19} \\ Z=\frac{6.24\times {10}^6\times 1.6\times {10}^{-19}\times 6.0\times {10}^{-15}}{1.6\times {10}^{-19}\times 2\times 1.6\times {10}^{-19}\times 9\times {10}^9}=13 \end{gathered}$$

Z = 13 for Aluminum