Some Nuclear Properties — Question 2

Question 2

Isotopic abundance determines the mean atomic mass

An idealized element contains isotopes of masses 62.0 u and 65.0 u. Its mean atomic mass is 63.2 u. Let f be the number fraction of the heavier isotope.

1. Find the abundance fractions.

2. Find the standard deviation of the mass of a randomly selected atom and distinguish it from the uncertainty in the mean of $10^6$ independent atoms.

Question 2 - Solution

Isotopic abundance determines the mean atomic mass

(a) Method and calculation

$$\bar m=(1-f)(62.0)+f(65.0)=62.0+3f.$$$$\boxed{f=(63.2-62.0)/3=.400,\quad1-f=.600}.$$

(b) Interpretation and further calculation

$$\boxed{\sigma_m^2=.6(62-63.2)^2+.4(65-63.2)^2=2.16\,\mathrm{u^2},\quad\sigma_m=1.470\,\mathrm u}.$$$$\boxed{\sigma_{\bar m}=\sigma_m/\sqrt{10^6}=.001470\,\mathrm u}.$$

Physical check

The isotope spread is a real variation among atoms. Averaging many independent atoms reduces uncertainty in the mean without narrowing that underlying distribution.