Notes: A sample of a radioactive substance decayed to 94.5% of its original amount after a year. a) What is the half-life of the substance? b) How long would...

Thursday, 4 February 2016

A sample of a radioactive substance decayed to 94.5% of its original amount after a year. a) What is the half-life of the substance? b) How long would...

Denote
the relative amount of the substance remaining after $\displaystyle{t}$ years as $\displaystyle{R}{\left({t}\right)}.$ We know that after
$\displaystyle{H}$ years (the half-life period) $\displaystyle\frac{{1}}{{2}}$ will remain: $\displaystyle{R}{\left({H}\right)}=\frac{{1}}{{2}}.$ Then after $\displaystyle{H}$ more years the
remaining amount will be $\displaystyle\frac{{1}}{{2}}$ of $\displaystyle\frac{{1}}{{2}},$ i.e. $\displaystyle\frac{{1}}{{4}},$ then $\displaystyle\frac{{1}}{{8}}$ and so on.

It
is a logarithmic decay function with the base 2, $\displaystyle{R}{\left({t}\right)}=\frac{{1}}{{{2}}^{{\frac{{t}}{{H}}}}}.$

a) First, we want to find the half-life H given that $\displaystyle{R}{\left({1}\right)}$
= 94.5%:

$\displaystyle\frac{{1}}{{{2}}^{{\frac{{1}}{{H}}}}}={94.5}%={0.945},{{2}}^{{\frac{{1}}{{H}}}}=\frac{{1}}{{0.945}}.$

Now take natural logarithm $\displaystyle{\ln{}}$ on both parts: $\displaystyle{\left(\frac{{1}}{{H}}\right)}{\ln{{\left({2}\right)}}}={\ln{{\left(\frac{{1}}{{0.945}}\right)}}},$ $\displaystyle{H}=$
ln(1/0.945) / ln(2) approx 12.25 (years).

We used $\displaystyle{\ln{}}$ because it is present
on most calculators.

b) Now we know $\displaystyle{R}{\left({t}\right)}\approx$
1/2^(t/12.25) $\displaystyle{\quad\text{and}\quad}{o}{u}{r}{g{{o}}}{a}{l}{i}{s}\to{f{\in}}{d}{f{{\quad\text{or}\quad}}}{w}{\hat{{t}}}{i}{t}{w}{i}{l}{l}{b}{e}{15}%.{I}{t}{i}{s}{t}{h}{e}{s}{a}{m}{e}{a}{s}{s}{o}{l}{v}{e}{t}{h}{e}$
equation $\displaystyle\frac{{1}}{{{2}}^{{\frac{{t}}{{12.25}}}}}={0.15}$ for t. It is not hard: