Estimating Time Since Formation of U 235 and U 238

[kel]

Homework Statement

Hi, I've worked this out (I think) but would appreciate someone xhecking over my method.

U 235 and U 238 found in ratio of 7.3E-3 to 1, but thought to have been produced equally (ie equal quantities)

half lives are 1.03E9 and 6.49E9 respectively.

Estimate time since formation

Homework Equations

Decay equation: N(t) = No e^(-wt) (used w for lambda here)

The Attempt at a Solution

As No is the same for both

Ln (N(t) / e^(-w1.t)) = Ln No

Ln N(t) + w1.t = Ln N(t) + w2.t

w1 = ln2 / 1.03E-9 = 6.729E-10
w2 = ln2 / 6.49E9 = 1.068E-10

Put these into above equ and re-arranging

ln (7.3E-3) - ln(1) = (t (6.729E-10)-(1.068E-10))

giving

t = (ln (7.3E-3) - ln(1)) / ((6.729E-10)-(1.068E-10))

which I work out to be 8.69E9 years.

I'm not quite sure if this is correct, but any advice would be appreciated.

Thanks

 

[variation]

Hi, your method is correct except a missed symbol in equation:
ln (7.3E-3) - ln(1) = t((6.729E-10)-(1.068E-10))
but it doesn't matter. If no calculation mistake, the answer is a correct one.

 

[kel]

If you mean that I have missed 't' out, then that is because both w1 and w2 are being multipled by t, so I factored it out

 

[nrqed]

Homework Statement

Hi, I've worked this out (I think) but would appreciate someone xhecking over my method.

U 235 and U 238 found in ratio of 7.3E-3 to 1, but thought to have been produced equally (ie equal quantities)

half lives are 1.03E9 and 6.49E9 respectively.

Estimate time since formation

Homework Equations

Decay equation: N(t) = No e^(-wt) (used w for lambda here)

The Attempt at a Solution

As No is the same for both

Ln (N(t) / e^(-w1.t)) = Ln No

Ln N(t) + w1.t = Ln N(t) + w2.t

w1 = ln2 / 1.03E-9 = 6.729E-10
w2 = ln2 / 6.49E9 = 1.068E-10

Put these into above equ and re-arranging

ln (7.3E-3) - ln(1) = (t (6.729E-10)-(1.068E-10))

giving

t = (ln (7.3E-3) - ln(1)) / ((6.729E-10)-(1.068E-10))

which I work out to be 8.69E9 years.

I'm not quite sure if this is correct, but any advice would be appreciated.

Thanks

Yes, it's right.

Notice you can always double check by plugging in the initial equation!

 

[Andrew Mason]

I am not clear on your units. The half life of U238 is 4.51x10^9 years. The half life of U235 is 7.04x10^8 years.

$$N/N0 = 1/2 =e^{\omega_{238}(4.51 x 10^9)}$$ so

$$\omega_{238} = \ln{.5}/(4.51 x 10^9) = -1.5x10^{-10}$$

Similarly:

$$\omega_{235} = \ln{.5}/(7.04 x 10^8) = -9.8x10^{-10}$$

So if they started out in the same proportion after a SuperNova, the SuperNova occurred about:

$$.0073 = e^{-\omega_{235}t}/e^{-\omega_{238}t} = e^{(\omega_{238}-\omega_{235})t}$$

$$t = \ln{.0073}/(\omega_{238}-\omega_{235}) = 4.92/(7.3x10^{-10}) = 6.74 x 10^9 \text{years}$$

AM

 

[Dick]

I am not clear on your units. The half life of U238 is 4.51x10^9 years. The half life of U235 is 7.04x10^8 years.

$$N/N0 = 1/2 =e^{\omega_{238}(4.51 x 10^9)}$$ so

$$\omega_{238} = \ln{.5}/(4.51 x 10^9) = -1.5x10^{-10}$$

Similarly:

$$\omega_{235} = \ln{.5}/(7.04 x 10^8) = -9.8x10^{-10}$$

So if they started out in the same proportion after a SuperNova, the SuperNova occurred about:

$$.0073 = e^{-\omega_{235}t}/e^{-\omega_{238}t} = e^{(\omega_{238}-\omega_{235})t}$$

$$t = \ln{.0073}/(\omega_{238}-\omega_{235}) = 4.92/(7.3x10^{-10}) = 6.74 x 10^9 \text{years}$$

AM

Good observation. It looks like the OP is being given inverse decay rates rather than half-lives. kel would be advised to carefully check the wording of the question.