OpenStudy (lgbasallote):

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: \[\LARGE \frac{(T_2-T_1)\ln 2}{\ln (\frac{M_1}{M_2})}\]

5 years ago
OpenStudy (lgbasallote):

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

5 years ago
OpenStudy (foolaroundmath):

\[ln(x_{o}) - ln(M_{1}) = kT_{1}\] \[ln(x_{o}) - ln(M_{2})= kT_{2}\] 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

5 years ago
OpenStudy (lgbasallote):

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

5 years ago
OpenStudy (foolaroundmath):

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

5 years ago
OpenStudy (lgbasallote):

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

5 years ago
OpenStudy (foolaroundmath):

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

5 years ago
OpenStudy (lgbasallote):

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

5 years ago
OpenStudy (foolaroundmath):

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

5 years ago
OpenStudy (lgbasallote):

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

5 years ago
OpenStudy (foolaroundmath):

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

5 years ago
OpenStudy (lgbasallote):

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

5 years ago
OpenStudy (foolaroundmath):

yeah

5 years ago
OpenStudy (lgbasallote):

why are they both at x?

5 years ago