Question

According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the universe in the "big bang." At present there are 137.7 atoms of 238U for each atom of 235U. Using the half-lives 4.51×109 years for 238U and 7.10×108 years for 235U, calculate the age of the universe.

Answer 1

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 $P(t)$ be the amount of $^{235}U$ and $Q(t)$ be the amount of $^{238}U$ after $t$ years.

Then, we obtain two differential equations

                               $\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q$

where $k_1$ and $k_2$ are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

                             $\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt$

Now, the variables are separated, $P$ and $Q$ appear only on the left, and $t$ appears only on the right, so that we can integrate both sides.

                         $\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt$

which yields

                      $\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2$,

where $c_1$ and $c_2$ are constants of integration.

By taking exponents, we obtain

                     $e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}$

Hence,

                            $P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}$,

where $C_1 := \pm e^{c_1}$ and $C_2 := \pm e^{c_2}$.

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

                                 $P(0) = Q(0) = C$

Substituting 0 for $P$ in the general solution gives

                         $C = P(0) = C_1 e^0 \implies C= C_1$

Similarly, we obtain $C = C_2$ and

                                $P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}$

The relation between the decay constant $k$ and the half-life is given by

                                            $\tau = \frac{\ln 2}{k}$

We can use this fact to determine the numeric values of the decay constants $k_1$ and $k_2$. Thus,

                     $4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}$

and

                     $7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}$

Therefore,

                              $P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}$

We have that

                                          $\frac{P(t)}{Q(t)} = 137.7$

Hence,

                                   $\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7$

Solving for $t$ yields $t \approx 6 \times 10^9$, which means that the age of the  universe is about 6 billion years.