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Deriving half thickness equation from exponential decay law?? (Medical Physics) watch

1. Attachment 724186
How do I derive the half thickness equation from the exponential decay law??
i.e.
Derive x1/2 = ln2/µ
From I=I0e^-µx

(the '1/2' and '0' are subscript)

Thank you!!
2. It's the thickness at which I = 0.5 I0

So, 0.5I0 = I0 e^-µx

Rearrange gives 0.5 = e^-µx

Then take ln of both sides, you'll need to rearrange again and remember that ln 0.5 = -ln2
3. I=I0e-µx

divide through by I0

I/I0=e-µx

and we're interested in the point x=x1/2 at which I=I0/2
1/0.5 = e-µx
2=e-µx

take logs of both sides
ln (2) = -µx
divide through by -µ

ln (2) /-µ=x
4. (Original post by phys981)
It's the thickness at which I = 0.5 I0

So, 0.5I0 = I0 e^-µx

Rearrange gives 0.5 = e^-µx

Then take ln of both sides, you'll need to rearrange again and remember that ln 0.5 = -ln2
(Original post by Joinedup)
I=I0e-µx

divide through by I0

I/I0=e-µx

and we're interested in the point x=x1/2 at which I=I0/2
1/0.5 = e-µx
2=e-µx

take logs of both sides
ln (2) = -µx
divide through by -µ

ln (2) /-µ=x
Thank you so much!!!

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