Discussion in 'MCAT Study Question Q&A' started by collegelife101, 05.17.14.

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1. ### collegelife101

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Hello everyone!

How do we know that the rate of decay is proportional to amount of radioisotope present? I would have thought it was constant. Is this just a fact that we need to memorize?

Thanks!

3. ### CzarcasmHakuna matata, no worries. 2+ Year Member

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Radioactive decay exhibits first order kinetics which means that the half-life of decay remains constant over time. So you're right there -- and that's a fact that's frequently emphasized and probably should be memorized. The actual rate though is dependent on the sample you have left. So for example if you start with 100g of radioactive sample which has a half life of say 10 seconds, then the rate of decay for the first decay process is essentially: amount decayed/time (ie. 50g/10s = 5g/s). Then when you have 50g remaining, the rate of decay for the second decay process is: 25g/10s = 2.5g/s. The rate changes based on how much product you have left. The actual half-life remains constant though.

You didn't actually have to memorize this if you reasoned it out. What is rate of decay? A rate is something per unit time, so rate of decay must be: what decayed/per time. But of course, the key part of answering this question would require you to realize half-life is constant.

4. ### Chrisz 2+ Year Member

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The decay rate is indeed first order, so the rate of decay is proportional to what is existing. Below is a derivation why half life is constant over the decay process
dA/dt=kA (k is negative)
ln(A)=kt+c
if t=0, c=ln(Ao)
ln(A)=kt+ln(Ao)
ln(A/A0)=kt
if decay reaches half life, A/A0=0.5
ln(0.5)/k=t
You see for first order, half life remains the same.
First order decay decreases exponentially, which also can be derived from here
ln(A)=kt+ln(A0)
A=Aoexp(kt)
exp(kt) is actually a term which can be considered as percent of population surviving at time t. In stats , it is called survival function, which can be derived from the probability density function, gamma distribution alpha=1, beta=-1/k (since we have defined k as negative in our original derivation).

5. ### CzarcasmHakuna matata, no worries. 2+ Year Member

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How much of this do you think OP will remember, lol.

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