Radioactive Decay: U234 & Th230 Half-Lives & Equal Atom Counts

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Homework Help Overview

The discussion revolves around the radioactive decay of Uranium-234 (U234) into Thorium-230 (Th230), focusing on the half-lives of both isotopes and the conditions under which the number of atoms of each type becomes equal. The original poster presents equations related to the decay process and seeks to determine the time at which the quantities of U234 and Th230 are equal, given initial conditions of 10^6 atoms of U234 and 0 atoms of Th230.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the mathematical relationships governing the decay process, including the use of exponential decay equations. Some participants question the validity of the original poster's approach and suggest alternative methods, such as writing differential equations for the decay rates. Others express confusion over the calculations and seek clarification on specific steps.

Discussion Status

The discussion is ongoing, with various participants offering insights and alternative perspectives on the problem. Some guidance has been provided regarding the formulation of equations, but there is no explicit consensus on the correct approach or resolution to the original poster's issue.

Contextual Notes

There are indications that assumptions regarding the decay process and the relationships between the decay constants may need to be revisited. The original poster's calculations have led to a negative logarithm, raising questions about the setup of the problem or the equations used.

Howlin
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Homework Statement


the half life of U234 is 245.5 years. It decays by α emmission to Th230 whos half life is 75.38. there are 10^6 atoms of U234 and 0 atoms of Th230 at t=0.
At what time will the number of atoms of each type be equal?

the half life of U234 is 245.5 years and half life of Th230 is 75.38
λa for U is 2.82e^-3 and λb for Th is 9.2e^-3

Homework Equations


Na = No*exp(-λut)
Nb = No*(λa/(λb-λa))*(exp(-λa*t)-exp(-λb*t))

The Attempt at a Solution


They Na and Nb are equal when Na = Nb

therefore:
No*exp(-λat) = No*(λa/(λb-λa))*(exp(-λa*t)-exp(-λb*t))
Divide by No and exp(-λat)
1 = (λa/(λb-λa))*(1-exp-(λa + λb)t)
Divide by (λb-λa)/a

1 - ((λb-λa)/λa) = exp-(λa + λb)t
getting t on its own
t = ln(1-((λb-λa)/λa)) / -(λb + λb)

t = ln(1-((9.2e-3-2.82e-3)/2.82e-3)) / -(2.82e-3 + 9.2e-3)
t = ln(1-(0.00638/2.82e-3)) / (- 0.01202)
t = ln(1-2.26) / (-0.01202)
t = ln(-1.26)/-.01202

here is where the problem is as you can not get the ln of a negitave number
Can anyone figure out where i have gone wrong?
or is there a problem with the question itself
 
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Hi Howlin!

Your solution was a bit messy for me to understand :-p so i couldn't figure out your error ...
however i can tell you how i would have solved it ,,,

You can write rate law of decay of U234 as
N = N_O e^{-\lambda_{U} t}

here N is number of atoms of U left

at the time when both U and Th have same number of atoms N\frac{N_O}{2} as half of U would give same number of Th atoms

now you can solve for t ...
 
If you go to
http://www-naweb.iaea.org/napc/ih/documents/global_cycle/vol I/cht_i_06.pdf

equation 6.5 and 6.18 are the two i used with the exception for 6.18 that i did not include the +etc in it

When the amount of the two materials are the same Na and Nb are equal the two equations are equal

so 6.5 = 6.18
you can rearrange the equations to get t on its own so you can find out when the time when the two substances have the same amount

(you still with me, what i done was just work out how i got to the final equation)

its just that when i put in my values for λ, the equation does not work
has this helped at all?
 
cupid.callin said:
You can write rate law of decay of U234 as
N = N_O e^{-\lambda_{U} t}

here N is number of atoms of U left

at the time when both U and Th have same number of atoms N\frac{N_O}{2} as half of U would give same number of Th atoms

now you can solve for t ...
That won't work here. At some time, half the U atoms are left. Question #1: Does that mean the remaining half of the original atoms are all Th? Answer: no. Question #2: Why not?

I recommend writing two differential equations, one for the number of Uranium atoms and another for the number of Thorium atoms:

\frac{dN_U}{dt} \ = \ ?
\frac{dN_{Th}}{dt} \ = \ ?
 
Curious3141 said:
I've previously posted a general solution to a problem on sequential decay. Might be helpful here. Here it is: https://www.physicsforums.com/showthread.php?t=119571

your Nb equation is one of the equations i am using and the other is the Na = Noexp(-λt)

We know that the amount of substance a remaining after time t is Na and the amount of substance b that is left after time t is Nb but i have to figure out after how long will both substances have the same number of particles?

This occurs when Na = Nb
I then made the two equtions equal to find out at what time this will occur
but when i rearrange the equations to get t on its own and work it out to get a value for t
this t value should be the time when both substances have the same atoms but i can not get this to work out for some reason and want to figure out why

curious 3141 can you work out an equation to find out at what time will two substances (A and B) will have the same amount of atoms if substance A decays into substance B and substance B decays into substance C

I have my answer in the first post and want to see if you get the same equation
 
Howlin said:
your Nb equation is one of the equations i am using and the other is the Na = Noexp(-λt)

We know that the amount of substance a remaining after time t is Na and the amount of substance b that is left after time t is Nb but i have to figure out after how long will both substances have the same number of particles?

This occurs when Na = Nb
I then made the two equtions equal to find out at what time this will occur
but when i rearrange the equations to get t on its own and work it out to get a value for t
this t value should be the time when both substances have the same atoms but i can not get this to work out for some reason and want to figure out why

curious 3141 can you work out an equation to find out at what time will two substances (A and B) will have the same amount of atoms if substance A decays into substance B and substance B decays into substance C

I have my answer in the first post and want to see if you get the same equation

OK, I see that. Your working is too messy to easily figure out and there are some obvious typos. Please format it into LaTex if you want us to review it.

That having been said, I believe the solution here is that there will never be a time when there are equal quantities of the two elements. For equality to exist, this condition must be met:

\lambda_A > \frac{1}{2}\lambda_B

or, equivalently, in terms of the respective half-lives:

T_A < 2 T_B

This condition is not met here. You will not get a real solution for the time.

Provided the condition is met, the time you want is \tau where:

\tau = \frac{1}{\lambda_A - \lambda_B}\ln(\frac{2\lambda_A - \lambda_B}{\lambda_A})
 
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At what time will hte number of atoms of both substances be equal if at t= there are 10^6 atoms of substance A and 0 atoms of substance B and substance A has a half life of 245.5 years and substance B has a half life of 75.38 years.

Here are the two equations for determining how many atoms of each substance is left at a particlar time


N_{A} = N_{O}e^{-λ_{A}t}

N_{B} = N_{O}\frac{λ_{A}}{λ_{A}-λ_{B}}(e^{-λ_{A}t} - e^{-λ_{B}t})

The time when both substances are equal is when

N_{A} = N_{B}

Therefore

N_{O}e^{-λ_{A}t} = N_{O}\frac{λ_{A}}{λ_{A}-λ_{B}}(e^{-λ_{A}t} - e^{-λ_{B}t})

Since N_{O} and e^{-λ_{A}t} are on both sides then we can gt rid of them and the equation becomes

1 = \frac{λ_{A}}{λ_{A}-λ_{B}}(1 - e^{-(λ_{A} +λ_{B})t})

We rearrange the equation to get the t value on its own on one side and the rest of the values on the other side

We divide by \frac{λ_{A}-λ_{B}}{λ_{A}}

1 - \frac{λ_{A}-λ_{B}}{λ_{A}}= e^{-(λ_{A} +λ_{B})t}
We get rid of the exp

ln(1 - \frac{λ_{A}-λ_{B}}{λ_{A}}) = -(λ_{A} +λ_{B})t

getting t on its own

t = [ln(1 - \frac{λ_{A}-λ_{B}}{λ_{A}}) ] / -(λ_{A} +λ_{B})

now the value we get for t should be the time when both substances have the same amount of atoms

But when i try and use this equation for two different questions (the one above and a different one where i know the answer) the equation do not give me the correct t value
 
Howlin said:
At what time will hte number of atoms of both substances be equal if at t= there are 10^6 atoms of substance A and 0 atoms of substance B and substance A has a half life of 245.5 years and substance B has a half life of 75.38 years.

Here are the two equations for determining how many atoms of each substance is left at a particlar time


N_{A} = N_{O}e^{-λ_{A}t}

N_{B} = N_{O}\frac{λ_{A}}{λ_{A}-λ_{B}}(e^{-λ_{A}t} - e^{-λ_{B}t})

The time when both substances are equal is when

N_{A} = N_{B}

Therefore

N_{O}e^{-λ_{A}t} = N_{O}\frac{λ_{A}}{λ_{A}-λ_{B}}(e^{-λ_{A}t} - e^{-λ_{B}t})

Since N_{O} and e^{-λ_{A}t} are on both sides then we can gt rid of them and the equation becomes

1 = \frac{λ_{A}}{λ_{A}-λ_{B}}(1 - e^{-(λ_{A} +λ_{B})t})

We rearrange the equation to get the t value on its own on one side and the rest of the values on the other side

We divide by \frac{λ_{A}-λ_{B}}{λ_{A}}

1 - \frac{λ_{A}-λ_{B}}{λ_{A}}= e^{-(λ_{A} +λ_{B})t}
We get rid of the exp

ln(1 - \frac{λ_{A}-λ_{B}}{λ_{A}}) = -(λ_{A} +λ_{B})t

getting t on its own

t = [ln(1 - \frac{λ_{A}-λ_{B}}{λ_{A}}) ] / -(λ_{A} +λ_{B})

now the value we get for t should be the time when both substances have the same amount of atoms

But when i try and use this equation for two different questions (the one above and a different one where i know the answer) the equation do not give me the correct t value

It's almost two in the AM here, so I can only take a cursory look at this. But right off the bat, there are two glaring errors:

The equation for N_B should read: N_{B} = N_{O}\frac{λ_{A}}{λ_{B}-λ_{A}}(e^{-λ_{A}t} - e^{-λ_{B}t})

Note the transposition of the lambdas. The equation you quoted in your first post was right (which also matches up with the equation I derived in my thread). You can see immediately (I hope) why this has to be the right form. If lambda A is higher than lambda B, the difference between the exponential terms would become negative. But the denominator of lambda B minus lambda A would also be negative, giving an overall positive value for N(B). Everything also works out to be positive in the other case if lambda B were to be higher than lambda A, so this is the correct form. In the version you started with, N(B) is always negative!

If you started wrong, the rest is going to be wrong. Nevertheless, here's another error, algebraic this time:

When you divide N_{O}e^{-λ_{A}t} = N_{O}\frac{λ_{A}}{λ_{A}-λ_{B}}(e^{-λ_{A}t} - e^{-λ_{B}t})

by N_{O} and e^{-λ_{A}t},

your result should be: 1 = \frac{λ_{A}}{λ_{A}-λ_{B}}(1 - e^{(λ_{A} -λ_{B})t})

See if you can see why. Here, I left the lambda terms reversed erroneously so as not to complicate your understanding, but when you redo it, make sure to start with the right equation.

There may be other errors, but this should be enough to start you off. If you're having trouble with the algebra, I suggest you be careful and work one step at a time.
 
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  • #10
Im sorry when i was typing the equation I made the mistake and the Nb one that u have is the one i have been using

I just noticed that when dividing by exp(-λt)
I forgot about the a negitave sign when dividing

thanks for all your help
It was a slide from a lecture and was basically copying him but then when i went step by step he made a mistake when dividing by e^{-λ_{A}t}
Thank you for all your help and happy holidays
 

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