How long does it take for the activity of the radionuclide to decrease by 50%?

[kirstlen]

Homework Statement

The activity (decays per second) of a certain radionuclide is observed to decrease by 87.5% in 30 hours.

What is the half-life of the radionuclide?

Homework Equations

(1) A(t) = A_{0}e^{-λt}

(2) T(1/2) = 0.693/λ

The Attempt at a Solution

I'm just stuck on which approach to use, do I need to use the above equations or would I simply be able to cross-multiply and divide like this:

\frac{0.875}{30 hrs} = \frac{0.50}{?} in which I find the half life to be 17.14 hours.

With the other approach I used equation 1 and solved for λ using a hypothetical A_{0} value of 100 so that A(t)=12.5 and found λ to be 0.0693 with t=30 hrs. Then I used the second equation to obtain T(1/2)=9.998 hours.

I'm just not sure which is the correct approach.

 

[gneill]

Your first method is assuming a linear relationship between rate of decay and time. This is not correct, since the rate of decay (number of decays per second) depends on the total number of atoms.

The second approach looks better, but could be improved. Have you studied calculus yet? If so, then you should know that the rate of decay (decays per second) will be given by the derivative w.r.t. time of your equation (1).

 

[kirstlen]

Oh! Thank you! This makes more sense now :)

 

[mjmontgo]

Hey so in regards to this how will taking the derivative help you? I also considered both her methods and realized the one assumed a linear relationship. Would you take the derivative and set it = 1/2? and even so you have more than one unknown variable...

 

[gneill]

The derivative gives you the rate of decay, decays per second, otherwise known as the activity. You're given a time span over which the activity changes by a given percentage. Set up a ratio.

 

[mjmontgo]

Ok sure, so the derivative of A(t)=A₀e^(-λt) is

A^'(t)= -λA₀e^(-λt) correct?

and you propose to set this equal to the 0.875/30 hours? Such that:
0.875/30hrs= -λA₀e^(-λt)
However you would have three unknowns still? I think I am just missing somthing small. But i just don't see how you can solve without (as mentioned above) finding the theoretical value for λ? I'm really not one for hand-holding I am just missing somthing fundamental or having a brain fart I think.

 

[gneill]

Ok sure, so the derivative of A(t)=A₀e^(-λt) is

A^'(t)= -λA₀e^(-λt) correct?

and you propose to set this equal to the 0.875/30 hours? Such that:
0.875/30hrs= -λA₀e^(-λt)

Okay, 0.875/30hrs makes no sense as a ratio; they're not the same things. One is time and the other a percentage.

The given percentage tells you how the rates are related at two different times. What are those times? Write the rate expression for both of them.

 

[mjmontgo]

Yes you're right that doesn't make sence... So for the rates as you said before i must use the derivative of the function
d/dt A(t)= -λA₀e^(-λt)

So:

A^'(30hr)= -λA₀e^(-30hrλ)=0.125

and the other? Is it just
A^'(t₀)= -λA₀e^(-λt₀)=1 ?

You'll have to forgive me if i don't immedietly see this. And if so do we consider t₀=0 ?

Then take the ratio of the equations?

 

[gneill]

Yes you're right that doesn't make sence... So for the rates as you said before i must use the derivative of the function
d/dt A(t)= -λA₀e^(-λt)

So:

A^'(30hr)= -λA₀e^(-30hrλ)=0.125

? Where does 0.125 come from?

You don't know (yet) what the rate is at t = 30hr. You only know that it's 87.5% of the initial rate (which you also don't know). So no numbers yet! Work with the symbols.

and the other? Is it just
A^'(t₀)= -λA₀e^(-λt₀)=1 ?

You'll have to forgive me if i don't immedietly see this. And if so do we consider t₀=0 ?

Yes. Counting from an initial arbitrary instant always begins at zero.

Then take the ratio of the equations?

Yes, because you know what that ratio should work out to be (that's where your percentage comes in).

 

[mjmontgo]

hahah oh wow, so you still have to solve for your lambda first. I thought there was going to be some way to directly solve for t_1/2 without first solving for lambda haha.


if that's the case i could of just as easily set up: 0.125A₀=A₀e^(-λt)
at t=30hrs haha this yields the same result giving λ=6.93x10^-2. Although i guess i just used intuition...and taking the ratio of the derivative is fundamentally what i did...weird haha. But thanks a lot for your help!Cheers!

This question just threw me through a loop weird they ask for t_1/2 initially and then λ as part b)...jeez.

 

[mjmontgo]

And just incase anyone else looks at this or needs help the correct equations are:

A^'(30)= -λA₀e^(-30λ)
A^'(t₀)= A^'(0) = -λA₀e⁽⁰⁾= -λA₀

(Because e⁰=1 of course)

And we know that A^'(30)/ A^'(0) = 0.125


Thus the ratio of A^'(30)/ A^'(0) = e^(-30λ)= 0.125

Taking ln of both sides and dividing by -30 yields λ= ln(0.125)/-30= 0.0692 (approx)
Using this lambda value you can set up your equation again (that is) A(t)= A₀e^(-λt) just setting A(t)= 1/2A₀. Now solve for your new time (half-life time).