Differention - rate of decay problem

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Arun Hallan
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How do you do this Q? ...

The rate of decay of a radioactive substance S is proportional to the amount remaining so that:

dx/dt = -kx

where x is the amount of S present at the time t and k is a positive constant. Given that x = a at
time t = 0 , show that, if the time taken for the amount of S to become (1/2)a is T then,

x = a(1/2) to the power of (t/T)

[i.e. x = a * (1/2) to the power of (t/T)

Please help, i've got my P3 exam on monday! For any1 else who does it, the Q is displayed better on
p101 of the Heinemann P3 Edexcel Book, Q65

Thanx in advance!
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Maxwell
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"Arun Hallan" <[email protected]> wrote in message
news:[email protected]...
[q1]> How do you do this Q? ...[/q1]
[q1]>[/q1]
[q1]>[/q1]
[q1]>[/q1]
[q1]> The rate of decay of a radioactive substance S is proportional to the amount remaining so that:[/q1]
[q1]>[/q1]
[q1]> dx/dt = -kx[/q1]
[q1]>[/q1]
[q1]>[/q1]
[q1]> where x is the amount of S present at the time t and k is a positive constant. Given that x = a[/q1]
[q1]> at time t = 0 , show that, if the time taken for the amount of S to become (1/2)a is T then,[/q1]
[q1]>[/q1]
[q1]> x = a(1/2) to the power of (t/T)[/q1]
[q1]>[/q1]
[q1]> [i.e. x = a * (1/2) to the power of (t/T)[/q1]
[q1]>[/q1]
[q1]>[/q1]
[q1]>[/q1]
[q1]>[/q1]
[q1]>[/q1]
[q1]> Please help, i've got my P3 exam on monday! For any1 else who does it, the Q is displayed better[/q1]
[q1]> on p101 of the Heinemann P3 Edexcel Book, Q65[/q1]
[q1]>[/q1]
[q1]>[/q1]
[q1]> Thanx in advance![/q1]

step (i): Seperate the variables ( i will use I for the integtation sign)

I 1/x dx = -k I dt

step (ii) Integrate

ln [x] = -kt + c (nb should be mod signs around x, but no prob here as x cant be neg anyway)

x = e^(-kt + c)

x = e^(-kt) . e ^c

x = Ae ^(-kt) ( where A=e^c)

when t = 0, x= a

a = A ( e^0)

A=a

x = ae^(-kt)

using the condition x=a/2 when t = T

m/2 = ae^(-kT)

1/2 = e^(-kT)

ln (1/2) = -kT

kT = ln 2 - ln 1

k = (ln 2) /T

therfore our general equation becomes

x = ae^(-ln(2) . t/T)

x= ae^(ln (1/2)) e^(t/T)

x= a.(1/2)e(t/T)

Max
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Arun Hallan
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Thanx Maxwell...

after all that i discovered that the book put the Q in the wrong place ... which led to me thinking
that i had to integrate it. The decay Q's are pretty easy, especially if u do physics
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