decomposition of Uranium 235?

Question

23592U is a naturally occurring radioisotope with a half-life of 7.1 x 108 yr. What percentage of this isotope that was present during the formation of Earth still remains? Assume that the Earth is 4.5 x 109 yr old and that no U-235 is created by decomposition of other elements.



a. 6.3%

b. 4.4%

c. 1.2%

d. 0.16%

e. 0.81%

I've worked through most of the math - k=8.76e-10
time = 4.5e9years - but how do I figure the percent remaining?

Answer 1

This is an interesting question.
This is similar to the Carbon decay you have asked earlier (I have to go back and change my computations. [see note below]).
In general radioactive decay can be modeled as
N(t)=N e^(-L t)
N(t) ni=number of atoms decayed fro time t
N - initial number of atoms
e - base of natural algorithm 2.71...
L -time constant
t –time
Note in your notation 4.5 x 109 = 4.5 E+9
**********************************
First let's find L
N(t)/N=e^(-Lt) take logs on both sides
ln(N(t)/N)= - Lt
since half lifefro Uranium 235 = 7.1E+8 years than (N(t)/N=1/2)
-L=-ln(2)/t
L=ln(2)/ 7.1 E+8 = 9.76 E-10
%=N(t)/N x 100 =e^(-Lt) x100=
%=e^(-9.76 E-10 x 4.5E+9 ) x 100=1.24 %
So the answer is (c)

Answer 2

% remaining = mass left / mass started with * 100%

say you start with x grams.
let y = amount remaining
n = number of half lifes that have gone by

the amount remaining (y) goes as this function ...

y = (1/2)^n * x

so that y/x * 100% = (1/2)^n x 100%
right?

in your case, n = number of half lifes gone by =
(4.5 x 10^9 yr ) / ( 7.1 x 10^8 yr) = 6.3 half lifes...

so that % remaining =

y/x * 100% = (1/2) ^ 6.3 * 100% = 1.3%

actually if you use 6.338 for n instead of rounding it to two sig figs, y/x*100% is 1.2% and the answer is "c"

Answer 3

At the start you can say 100% remains.