# Thread: can u solve the problem???

1. ## can u solve the problem???

radium decomposes at a rate propotional to the amount present.if p percent of the original amount disappears in l years how much will remain at the end of 2l years?

2. Let m be the percentage mass of radium present at time t.

Let $\displaystyle m_0$ be the mass present at time $\displaystyle t=0$.

$\displaystyle \frac{dm}{dt} \propto m$

$\displaystyle \frac{dm}{dt}=km$ where k is a real constant (k is actually negative since this is "exponential decay". You can write k=-b or something to show that you know it is negative. I just left it because it doesn't change the overall answer at the end.)

Integrating gives:

$\displaystyle \int \frac{1}{m} dm=\int k dt$

$\displaystyle ln m=kt + C$

$\displaystyle m=e^{kt+C}$

$\displaystyle m=Ae^{kt}$ where $\displaystyle A=e^C$

using the intial conditions:

$\displaystyle m=m_0 e^{kt}$

when t=I years:

$\displaystyle m_I=m=m_0e^{Ik}$

At t=2I years:

$\displaystyle m=m_0e^{2Ik}$

$\displaystyle m=m_0(e^{Ik})^2$

$\displaystyle m=m_0(e^{Ik})(e^{Ik})$

$\displaystyle m=m_I(e^{Ik})$

and I think we're done! =D

P.S: I did change notation. I used m instead of p. If you like you can work it through again with m=p or just write this at the end.

Radium decomposes at a rate propotional to the amount present.
If $\displaystyle p$ percent of the original amount disappears in $\displaystyle L$ years.
how much will remain at the end of $\displaystyle 2L$ years?
Let $\displaystyle Q$ = quantity of radium at time $\displaystyle t.$

We have: .$\displaystyle \frac{dQ}{dt} \:=\:-kQ \quad\Rightarrow\quad \frac{dQ}{Q} \:=\:-kt$

Integrate: .$\displaystyle \ln(Q) \:=\:-kt + C \quad\Rightarrow\quad Q \:=\:e^{-kt+c} \quad\Rightarrow\quad Q \:=\:Ce^{-kt}$

When $\displaystyle t = 0,\;Q = Q_o$ (initial amount)
We have: .$\displaystyle Q_o \:=\:Ce^0 \quad\Rightarrow\quad C \:=\:Q_o$

The function (so far) is: .$\displaystyle Q \;=\;Q_o\,e^{-kt}$

When $\displaystyle t = L,\;Q \:=\:(1-p)Q_o$

So we have: .$\displaystyle (1-p)Q_o \;=\;Q_o\,e^{-kL} \quad\Rightarrow\quad e^{-kL} \:=\:1-p \quad\Rightarrow\quad -kL \:=\:\ln(1-p)$

. . Hence: .$\displaystyle k \:=\:-\frac{\ln(1-p)}{L}$

The function is: .$\displaystyle Q \;=\;Q_o\,e^{\text{-}\left[\text{-}\frac{ln(1-p)}{L}\right]t} \quad\Rightarrow\quad Q \;=\;Q_o\,e^{\frac{\ln(1-p)}{L}\,t}$

When $\displaystyle t = 2L\!:\;\;Q \;=\;Q_o\,e^{\frac{\ln(1-p)}{L}\cdot2L} \;=\;Q_o\,e^{2\ln(1-p)} \;=\;Q_o\,e^{\ln(1-p)^2}$

We have: .$\displaystyle Q \;=\;Q_o(1-p)^2$

Therefore: .$\displaystyle \frac{Q}{Q_o} \;=\;(1-p)^2\quad _{\Leftarrow\;\;\text{Percent of radium after 2L years}}$

. .$\displaystyle \overbrace{_{\text{Fraction of radium after 2L years}}}^{\Uparrow}$