3. A radio active material, such as the isotope thorium-234, disintegrates at a

Question

3. A radio active material, such as the isotope thorium-234, disintegrates at a rate proportional to the amount currently present. If gt) is the amount present at time t, then dQ/dtQ where r > 0 is the decay rate. -TQ, (a) If 100 mg of thorium-234 decays to 82.04 mg in 1 week, determine the decay rate (b) The half-life of a radioactive material is the time required for an amount of the material to decay to one-half its original value. Show that the half-life ? and the decay rate r satis the equation rT In 2.

Explanation / Answer

a) Given dQ/dt = - rQ

So separting the variable and integrating

dQ/Q = - r dt

On intergrating the above equation we get

ln(Q) = - rt + c .. ... Eq1

C= integrating constant

At time t =0

Q = Qo since no material is decayed (Qo= intial quantity of material)

So puting the value t=0 and Q=Qo in Eq1 we get

ln(Qo) = 0+ c

So c = ln(Qo)

So the final solution becomes

ln(Q) = - rt +ln(Qo)

On rearranging the above equation

ln(Q/Qo) = - rt eq2 .. This is the decaying equation of the material

So in the given question at time t=1 week the Q= 82.04 mg

And Q = 100

So putting all the values jn the eq 2

We get ln(82.04/100) = - r*1

r = 0.197 mg per week

b) At half life Q will become Qo/2 so putting the eq2 we get

ln(Qo/2*Qo) = - rT ,

T =halflife time, i am representing it with 'T'

Solving above eq and cancelling Qo we will get

-ln(1/2) =-rT

Cancelling -

We get

ln(1/2) = rT

Hence proved

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