Closest Approach Estimate (AQA A Level Physics): Revision Note

The first artificially produced isotope, phosphorus-30 $\left({}_{15}\mathrm{P}\right)$ was formed by bombarding an aluminium-27 isotope $\left({}_{13}\mathrm{Al}\right)$ with an α-particle.

For the reaction to take place, the α-particle must come within a distance $r$ from the centre of the aluminium nucleus.

Calculate the distance $r$ for an α-particle accelerated to a speed of 2.55 × 10⁷ m s^–1.

Answer:

Step 1: List the known quantities

• Mass of α-particle, $m=4u=4\left(1.66\times {10}^{-27}\right)\mathrm{kg}$

• Speed of the α-particle, $v=2.55\times {10}^{-7}\mathrm{m}{\mathrm{s}}^{-1}$

• Charge of an α-particle, $q=2e=2\left(1.6\times {10}^{-19}\right)\mathrm{C}$

• Proton number of aluminium, $Z=13$

• Charge of an aluminium nucleus, $Q=13e=13\left(1.6\times {10}^{-19}\right)\mathrm{C}$

• Permittivity of free space, ${\epsilon }_0=8.85\times {10}^{-12}\mathrm{F}{\mathrm{m}}^{-1}$

Step 2: Write down the equations for kinetic energy and electric potential energy

$E_k=E_p=\frac{1}{2}m{v}^2=\frac{Qq}{4\mathrm{\pi }{\epsilon }_{0}r}$

$\frac{1}{2}m{v}^2=\frac{2Z{e}^2}{4\mathrm{\pi }{\epsilon }_{0}r}$

Step 3: Rearrange for distance, r

$r=\frac{Z{e}^2}{\mathrm{\pi }{\epsilon }_{0}m{v}^2}$

Step 4: Calculate the distance, r

$r=\frac{13{\left(1.6\times {10}^{-19}\right)}^2}{\mathrm{\pi }\left(8.85\times {10}^{-12}\right)\left(4\right)\left(1.66\times {10}^{-27}\right){\left(2.55\times {10}^7\right)}^2}$

$r=2.77\times {10}^{-15}\mathrm{m}$

Make sure you're comfortable with the calculations involved with the alpha particle closest approach method, as this is a common exam question. You will be expected to remember that the charge of an α-particle is the charge of 2 protons (2 × the charge of an electron)