Question

The 238U isotope has a half-life T238 = 4.5x10^9 years and the 235U has
T235 = 7.0x10^8 years.
N238(t) is the number of 238U nuclei at time t and
N235(t) is the corresponding number for 235U.
The relative abundance r(t)
is defined as r(t) = N235(t)/N238(t). At present, r = 0.0072.

Estimate the relative abundance of these two isotopes 109 years ago. You might use the following approximations: e^x ~ 1 +x for small x, e ~ 2.7 and ln 2 ~ 0.7.

Answer 1

sorry, trying to remember this :P 1 sec...

Answer 2

I think this is right....

\[
\frac{dN}{dt} = -\lambda N \ \ \ \textrm{Rate of change of substance N equal to negative constant}
\\ \
\\ \int_{N_o}^{N}\frac{dN}{N} = - \int_0^t\lambda dt
\\ \
\\ \
\\ ln(N) - ln(N_o) = - \lambda t
\\ \
\\ ln{ \frac{N}{N_o}} = \lambda t
\\ \
\\ \
\\ N_{(t)} = N_oe^{-\lambda t}
\\ \
\\ \textrm{Now we want to find the decay constant } \lambda
\\ \textrm{We can do this using our half life times}
\\ \
\\ t_{\small 238} = 4.5x10^9 \ years
\\ t_{\small 235} = 7x10_8 \ years
\\ \
\\ N_{t_{\small {1/2}}}^{\small 238} =\frac{1}{2} N_o^{\small 238} =N_o^{\small 238} e ^ {-\lambda_{1} t_{\small {238}}}
\\ \
\\ \lambda_{1} t_{ \small{238}} = \ln 2
\\ \
\\ \lambda_1 = \frac{ \ln 2}{t_{238}}
\\ \
\\ \
\\ \textrm{Similarly}
\\ \
\\ \lambda_2 = \frac{\ln 2}{t_{235}}
\]
Follow so far?

Answer 3

N(t)=Noe^−λt
this is equivalent to the Ae^-kt right?

I get the two lambda equations

Answer 4

Yep yep ^^

Answer 5

\[
N_{(t)}^{238} = N_o^{238}e^{-\lambda_1 t}
\\ \
\\ N_{(t)}^{235} = N_o^{235}e^{-\lambda_2 t}
\\ \
\\ \
r_{(t)} = \frac{N_{(t)}^{235}}{N_{(t)}^{238}} = \frac{N_o^{235}e^{-\lambda_2 t}}{N_o^{238}e^{-\lambda_1 t}} = \frac{N_o^{235}}{N_o^{238}}e^{(\lambda_1-\lambda_2)t}
\\ \
\\ \
\\ r_{(0)} = \frac{N_o^{235}}{N_o^{238}} e^0 = .0072
\\ \
\\ r_{(t)} = r_o e^{(\lambda_2-\lambda_1)t}
\\ \
\\ \textrm{the exponent should be really small, so}
\\ \
\\ r_{(t)} = r_o(\lambda_1 - \lambda_2)t
\]
Then plug in -109 years. Should be good to go I think ^_^

Answer 6

nope nope nope, that last one is wrong. sorry

Answer 7

\[ r_{(t)} = r_o(1+(\lambda_1 - \lambda_2)t)\]

Answer 8

I got 7/6250 - which does not seem like a correct answer. Proportion of U235 to U238 should be higher 10^9 years back I think

Answer 9

Unless the ratio is 81/1750. i think i mixed up N(o) and N(t)

Answer 10

^^