Discovering the Nucleus — Question 6

Question 6

Foil comparisons isolate the square of nuclear charge

Two thin foils are measured at identical beam energy, angle, incident fluence and detector geometry. Their areal nuclear densities are $n_{t1}$ and $n_{t2}=.500n_{t1}$. Foil 1 has $Z_1=30$; the background-corrected count ratio is $N_2/N_1=2.00$. Assume Rutherford scattering.

1. Infer the charge number of foil 2.

2. If instead the second beam energy were 1.20 times the first with the same measured ratio, infer the corresponding charge number.

Question 6 - Solution

Foil comparisons isolate the square of nuclear charge

(a) Method and calculation

$$N\propto n_tZ^2/E^2\Rightarrow\frac{N_2}{N_1}=\frac{n_{t2}}{n_{t1}}\left(\frac{Z_2}{Z_1}\right)^2.$$$$\boxed{Z_2=30\sqrt{2.00/.500}=60}.$$

(b) Interpretation and further calculation

$$Z_2=Z_1(E_2/E_1)\sqrt{(N_2/N_1)(n_{t1}/n_{t2})}.$$$$\boxed{Z_2=30(1.20)(2)=72}.$$

Physical check

Comparing mass thicknesses without converting to numbers of target nuclei would bias the inferred charge. The energy factor is equally important.