Find half-life from an atomic ratio?

[doanta]

Homework Statement

An equilibrium ore mixture has an atomic ratio for [tex]\frac{U-235}{Pa-231} = 3.04x10^{6}[/tex]

Find the half-life of U-235 from the data. The known half-life for Pa-231 is [tex]3.28x10^{4}[/tex] years

Homework Equations

For secular equilibrium
[tex]\frac{\lambda_{2} N_{2}}{\lambda_{1} N_{1}} = 1
[/tex]

[tex]\lambda = \frac{Ln(2)}{Half-Life}[/tex]

The Attempt at a Solution

[tex]\lambda_{Pa-231} = \frac{Ln(2)}{3.28x10^{4}} = 2.113x10^{-5}y^{-1}[/tex]

[tex]\frac{2.113x10^{-5}y^{-1}}{\lambda_{1}} = 3.04x10^{6}[/tex]

[tex]\lambda_{1} = 6.951x10^{-12}[/tex]

[tex]T_{1/2_{U}} = \frac{Ln(2)}{6.951x10^{-12}}[/tex]

[tex]T_{1/2_{U}} = 9.97x10^{10}yrs[/tex]

Am I taking the right approach? I know the half life of U-235 is only [tex]7.04x10^{8}[/tex] and my answer is pretty far off.

 

[anathema]

Your approach is correct, but there seems to be an error in your calculation. When you calculated the value for \lambda_{Pa-231}, you used the half-life of 3.28x10^{4} years, but then in the next step, you used the value of 2.113x10^{-5}y^{-1} as \lambda_{Pa-231}. This value is actually \lambda_{U-235}.

The correct calculation would be:

\frac{\lambda_{U-235} N_{U-235}}{\lambda_{Pa-231} N_{Pa-231}} = 1

\frac{2.113x10^{-5}y^{-1} N_{U-235}}{2.113x10^{-5}y^{-1} N_{Pa-231}} = 1

N_{U-235} = 3.04x10^{6} N_{Pa-231}

Using the known half-life for Pa-231, \lambda_{Pa-231} = \frac{Ln(2)}{3.28x10^{4}} = 2.113x10^{-5}y^{-1}, we can solve for N_{Pa-231}:

N_{Pa-231} = \frac{3.04x10^{6}}{2.113x10^{-5}} = 1.44x10^{11}

Now, we can use the value for N_{Pa-231} to solve for the half-life of U-235:

\lambda_{U-235} = \frac{Ln(2)}{T_{1/2_{U}}}

2.113x10^{-5}y^{-1} = \frac{Ln(2)}{T_{1/2_{U}}}

T_{1/2_{U}} = \frac{Ln(2)}{2.113x10^{-5}y^{-1}}

T_{1/2_{U}} = 3.28x10^{4} years

This is the same value as the known half-life for Pa-231, so it seems that there may have been an error in the given data or in the calculation of the atomic ratio.

In summary, your approach is correct, but there was a small error in the calculation. Keep in mind to always double check your values and units to ensure accuracy in