Answer:
6 billion years.
Step-by-step explanation:
According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let
be the amount of
and
be the amount of
after
years.
Then, we obtain two differential equations
![\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%20%3D%20-k_1P%20%5Cquad%20%5Cfrac%7BdQ%7D%7Bdt%7D%20%3D%20-k_2Q)
where
and
are proportionality constants and the minus signs denotes decay.
Rearranging terms in the equations gives
![\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7BP%7D%20%3D%20-k_1dt%20%5Cquad%20%5Cfrac%7BdQ%7D%7BQ%7D%20%3D%20-k_2dt)
Now, the variables are separated,
and
appear only on the left, and
appears only on the right, so that we can integrate both sides.
![\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7BdP%7D%7BP%7D%20%3D%20-k_1%20%5Cint%20dt%20%5Cquad%20%5Cint%20%5Cfrac%7BdQ%7D%7BQ%7D%20%3D%20-k_2%5Cint%20dt)
which yields
,
where
and
are constants of integration.
By taking exponents, we obtain
![e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}](https://tex.z-dn.net/?f=e%5E%7B%5Cln%20%7CP%7C%7D%20%3D%20e%5E%7B-k_1t%20%2B%20c_1%7D%20%20%5Cquad%20e%5E%7B%5Cln%20%7CQ%7C%7D%20%3D%20e%5E%7B-k_12t%20%2B%20c_2%7D)
Hence,
,
where
and
.
Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition
![P(0) = Q(0) = C](https://tex.z-dn.net/?f=P%280%29%20%3D%20Q%280%29%20%3D%20C)
Substituting 0 for
in the general solution gives
![C = P(0) = C_1 e^0 \implies C= C_1](https://tex.z-dn.net/?f=C%20%3D%20P%280%29%20%3D%20C_1%20e%5E0%20%5Cimplies%20C%3D%20C_1)
Similarly, we obtain
and
![P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}](https://tex.z-dn.net/?f=P%20%20%3D%20Ce%5E%7B-k_1t%7D%20%5Cquad%20Q%20%20%3D%20Ce%5E%7B-k_2t%7D)
The relation between the decay constant
and the half-life is given by
![\tau = \frac{\ln 2}{k}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cfrac%7B%5Cln%202%7D%7Bk%7D)
We can use this fact to determine the numeric values of the decay constants
and
. Thus,
![4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}](https://tex.z-dn.net/?f=4.51%20%5Ctimes%2010%5E9%20%3D%20%5Cfrac%7B%5Cln%202%7D%7Bk_1%7D%20%5Cimplies%20k_1%20%3D%20%5Cfrac%7B%5Cln%202%7D%7B4.51%20%5Ctimes%2010%5E9%7D)
and
![7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}](https://tex.z-dn.net/?f=7.10%20%5Ctimes%2010%5E8%20%3D%20%5Cfrac%7B%5Cln%202%7D%7Bk_2%7D%20%5Cimplies%20k_2%20%3D%20%5Cfrac%7B%5Cln%202%7D%7B7.10%20%5Ctimes%2010%5E8%7D)
Therefore,
![P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}](https://tex.z-dn.net/?f=P%20%20%3D%20Ce%5E%7B-%5Cfrac%7B%5Cln%202%7D%7B4.51%20%5Ctimes%2010%5E9%7Dt%7D%20%5Cquad%20Q%20%20%3D%20Ce%5E%7B-k_2%20%3D%20%5Cfrac%7B%5Cln%202%7D%7B7.10%20%5Ctimes%2010%5E8%7Dt%7D)
We have that
![\frac{P(t)}{Q(t)} = 137.7](https://tex.z-dn.net/?f=%5Cfrac%7BP%28t%29%7D%7BQ%28t%29%7D%20%3D%20137.7)
Hence,
![\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7](https://tex.z-dn.net/?f=%5Cfrac%7BCe%5E%7B-%5Cfrac%7B%5Cln%202%7D%7B4.51%20%5Ctimes%2010%5E9%7Dt%7D%20%7D%7BCe%5E%7B-k_2%20%3D%20%5Cfrac%7B%5Cln%202%7D%7B7.10%20%5Ctimes%2010%5E8%7Dt%7D%7D%20%3D%20137.7)
Solving for
yields
, which means that the age of the universe is about 6 billion years.