The half-life for the process 238U→206Pb is 4.5×109yr. A mineral sample contains 67.0 mg of 238U and 16.5 mg of 206Pb.
What is the age of the mineral? Express your answer using two significant figures.
To find the age of the mineral, we can use the concept of half-life. The half-life of a radioactive substance is the time it takes for half of the substance to decay.
In this case, the half-life for the process 238U → 206Pb is given as 4.5×10^9 years.
First, we need to calculate the number of half-lives that have occurred since the mineral sample was formed. We can do this by dividing the total amount of 206Pb in the sample (16.5 mg) by the amount of 206Pb that was initially present in the sample, assuming it was all produced from the decay of 238U:
Number of half-lives = (amount of 206Pb in the sample) / (amount of 206Pb produced from 238U decay)
Since the decay process is described by the equation 238U → 206Pb, the stoichiometry of the reaction tells us that the amount of 206Pb produced is equal to the amount of 238U decayed. So, we can equate the amount of 206Pb in the sample (16.5 mg) to the amount of 238U that initially decayed:
16.5 mg = (amount of 238U initially present) - (amount of 238U remaining)
We can rearrange this equation to solve for the amount of 238U initially present:
(amount of 238U initially present) = (amount of 238U remaining) + 16.5 mg
Next, we need to calculate the remaining amount of 238U in the sample. As mentioned earlier, the half-life of 238U is 4.5×10^9 years. We can use the formula for radioactive decay to calculate the remaining amount of 238U:
(amount of 238U remaining) = (amount of 238U initially present) × (1/2)^(number of half-lives)
Substituting the previously calculated value for (amount of 238U initially present), we can solve for (amount of 238U remaining).
Finally, we need to calculate the age of the mineral, which is the number of half-lifes multiplied by the half-life of the process. We can use this equation:
Age = (number of half-lives) × (half-life)
Substituting the previously calculated values for the number of half-lives and the half-life, we can calculate the age of the mineral sample.
It's important to note that we express the answer using two significant figures as requested.
so you have 67.0/83.5 = 0.8024 of the original U238
You need to solve
(1/2)^(t/(4.5*10^9)) = 0.8024
t = 1.42923*10^9 years
That's about 1/3 of a half-life
ln (No/N) = kt where
k = 0.693/4.5E9 = ? Solve for k.
The sample started out as x mg U 238 and 0 mg Pb 206. Pb 206 now is 16.5 mg. What amount of U 238 converted to Pb 206 during the sample's life? That is 16.5 mg x (238/206) = 19.1 mg U. Starting amount of U 238 was 19.1 mg + 67.0 mg = 86.1mg. Plug all of this into
ln (No/N) = kt
ln(86.1/67.0) = kt. k is from above but I didn't calculate it.
Solve for t in years and round as per the problem to two significant figures.
Post your work if you get stuck.