Uranium disintegrates at a rate proportional to the amount present at any time. If $M_1$ and $M_2$ grams of uranium are present at $T_1$ and $T_2$ respectively, show that the half-life of uranium is:
$$\LARGE \frac{(T_2-T_1)\ln 2}{\ln (\frac{M_1}{M_2})}$$

Question

Uranium disintegrates at a rate proportional to the amount present at any time. If $M_1$ and $M_2$ grams of uranium are present at $T_1$ and $T_2$ respectively, show that the half-life of uranium is:
$$\LARGE \frac{(T_2-T_1)\ln 2}{\ln (\frac{M_1}{M_2})}$$

lgbasallote · Answer

the formula for decay is $$\LARGE \ln (\frac{x_o}{x}) = kt$$
how can i use that here?

foolaroundmath · Answer

$$ln(x_{o}) - ln(M_{1}) = kT_{1}$$ $$ln(x_{o}) - ln(M_{2})= kT_{2}$$ subtracting these two we get  $$ln\frac{M_{2}}{M_{1}} = k(T_{1}-T_{2})$$ use the above expression to substitute for k.
At time t=0, the mass = x_o and at time T the mass is x_o/2 (the definition of half life). Doing as above, we get
$$ln\frac{1}{2} = k(0-T_{0.5}) \Rightarrow T_{0.5} = \frac{ln2}{k}$$Substitute the value of k obtained above

lgbasallote · Answer

uhmm wait you lost me from the first statement..where did that come from?

foolaroundmath · Answer

ln(a/b) = ln(a) - ln(b)

lgbasallote · Answer

but why ln x0 - ln m1 = kt1? where did that come from?

foolaroundmath · Answer

your formula for decay. $$ln\frac{x_{o}}{x} = ln(x_{o}) - ln(x) $$

lgbasallote · Answer

okay...i get this part..then..what did you do?

foolaroundmath · Answer

set x = M1 at T1 and x = M2 at T2.. then subtract the two equations

lgbasallote · Answer

wait why? im sure the substitution cant be pulled out of nowhere..

foolaroundmath · Answer

you are given that the mass at time T1 is M1 and at time T2 is M2 !

lgbasallote · Answer

wait..what's the substitution again? x = m1 at t1 then x = m2 at t2??

foolaroundmath · Answer

yeah

lgbasallote · Answer

why are they both at x?