Respuesta :
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 [tex]P(t)[/tex] be the amount of [tex]^{235}U[/tex] and [tex]Q(t)[/tex] be the amount of [tex]^{238}U[/tex] after [tex]t[/tex] years.
Then, we obtain two differential equations
                [tex]\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q[/tex]
where [tex]k_1[/tex] and [tex]k_2[/tex] are proportionality constants and the minus signs denotes decay.
Rearranging terms in the equations gives
               [tex]\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt[/tex]
Now, the variables are separated, [tex]P[/tex] and [tex]Q[/tex] appear only on the left, and [tex]t[/tex] appears only on the right, so that we can integrate both sides.
             [tex]\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt[/tex]
which yields
           [tex]\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2[/tex],
where [tex]c_1[/tex] and [tex]c_2[/tex] are constants of integration.
By taking exponents, we obtain
           [tex]e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}[/tex]
Hence,
              [tex]P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}[/tex],
where [tex]C_1 := \pm e^{c_1}[/tex] and [tex]C_2 := \pm e^{c_2}[/tex].
Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition
                 [tex]P(0) = Q(0) = C[/tex]
Substituting 0 for [tex]P[/tex] in the general solution gives
             [tex]C = P(0) = C_1 e^0 \implies C= C_1[/tex]
Similarly, we obtain [tex]C = C_2[/tex] and
                [tex]P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}[/tex]
The relation between the decay constant [tex]k[/tex] and the half-life is given by
                      [tex]\tau = \frac{\ln 2}{k}[/tex]
We can use this fact to determine the numeric values of the decay constants [tex]k_1[/tex] and [tex]k_2[/tex]. Thus,
           [tex]4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}[/tex]
and
           [tex]7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}[/tex]
Therefore,
               [tex]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}[/tex]
We have that
                     [tex]\frac{P(t)}{Q(t)} = 137.7[/tex]
Hence,
                  [tex]\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[/tex]
Solving for [tex]t[/tex] yields [tex]t \approx 6 \times 10^9[/tex], which means that the age of the  universe is about 6 billion years.
