The half-life for the process 238U 206Pb is 4.5 109 yr. A mineral sample contains 45.9 mg of 238U and 16.8 mg of 206Pb. What is the age of the mineral?
i used the equation: n(T)=n(O)/2^(t/halflife) and i got 6.53e9 but it is wrong. what am i doing wrong?
I suspect you didn't change the Pb to U.
16.8 mg Pb x (238/206) = 19.4 mg U
19.4 + original U = 19.4 + 45.9 = 65.3; therefore, the original amount of U238 = 65.3.
Now k = 0.693/t1/2 = 0.693/4l.5E9.
Then ln(No/N) = kt
No = 65.3
N = 45.9
k from above.
Solve for t in years.
thanks! so k=1.669e-11
so ln(65.3/45.9)=1.669e-11(t)
so does t=2.112e10 years?
I entered the asnwer I got and it is wrong.
To determine the age of the mineral using the given equation, you are on the right track. However, there seems to be a mistake in your calculation. Let's break down the process to find where the error may have occurred.
The equation you are using is:
n(T) = n(O) / 2^(t/halflife)
Where:
n(T) is the remaining amount of parent isotope (238U in this case) after time T
n(O) is the initial amount of parent isotope
t is the age of the mineral we are trying to find
halflife is the half-life of the parent isotope (4.5 x 10^9 years)
The given information is:
- n(O) = 45.9 mg (initial amount of 238U)
- n(T) = 16.8 mg (amount of 206Pb, which is the daughter isotope)
- halflife = 4.5 x 10^9 years
We need to solve for t, the age of the mineral.
Rearranging the equation, we have:
2^(t/halflife) = n(O) / n(T)
Taking the logarithm of both sides, we get:
log₂(2^(t/halflife)) = log₂(n(O) / n(T))
Using the logarithm property log₂(a^b) = b * log₂(a), we can simplify:
(t/halflife) * log₂(2) = log₂(n(O) / n(T))
Since log₂(2) equals 1, the equation further simplifies to:
t/halflife = log₂(n(O) / n(T))
Finally, isolating t, we obtain:
t = halflife * log₂(n(O) / n(T))
Now let's substitute the given values:
t = (4.5 x 10^9 years) * log₂(45.9 mg / 16.8 mg)
Using this formula and performing the calculation, the correct result for the age of the mineral should be approximately 6.24 x 10^9 years.
Make sure to double-check your calculations, particularly the logarithm part, to identify where the error occurred.