Solving Radioactive Decay: Homework Equations and Attempt at Solution

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elemis
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Homework Statement



radioactive.png


Homework Equations



N=N0e-λt

The Attempt at a Solution



I can do all of part (a) but I'm stuck on part (b).

I don't even know where to begin. I just have no idea what to make of this question.

Could someone prod me in the right direction ?
 
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What's the expression for the activity of a sample over time, given its half-life and initial activity value?
 
gneill said:
What's the expression for the activity of a sample over time, given its half-life and initial activity value?

I've written down that expression in the original post.
 
elemis said:
I've written down that expression in the original post.

Not quite, you've written down the amount of material remaining as a function of time. The rate of decay is measured in Becquerels (Bq) and represents the instantaneous number of decays per second. It has a similar form to the equation you wrote...
 
gneill said:
Not quite, you've written down the amount of material remaining as a function of time. The rate of decay is measured in Becquerels (Bq) and represents the instantaneous number of decays per second. It has a similar form to the equation you wrote...

Its basically the same equation except N is subbed for A and N0 for A0
 
elemis said:
Its basically the same equation except N is subbed for A and N0 for A0
Correct. And you're given a value for Ao are you not?

What would constitute a "current" value for A that would represent a maximum deviation of 10% from Ao? Keep in mind that Ao has an initial uncertainty of 2%.
 
gneill said:
Correct. And you're given a value for Ao are you not?

What would constitute a "current" value for A that would represent a maximum deviation of 10% from Ao? Keep in mind that Ao has an initial uncertainty of 2%.

So A0 = 3.5*105

3.78*105 ? A 2% uncertainty is built into A0 so an 8% error must originate from the A ?

So 0.08*3.5*10^5 + 3.5*10^5 = above answer ?
 
You're thinking along the right lines... but keep in mind that the Activity must decrease over time. So the new value of A should be smaller than Ao.