Question

(a) consider the initial-value problem a=ka,a(0)=ao as t as the model for the decay of a radioactive substance. show that, in general, the half-life t of the substance is t = -(ln 2)/k. (b) if a radioactive substance has the half-life t given in part (a), how long will it take an initial amount aoof the substance to decay to 01/6ad?

Answer 1

The initial-value problem dA/dt = kA, A(0) = A₀ is used to represent the radioactive decay. The radioactive substance having half-life T= -(ln 2)/k,will take 2.5 T for the substance to decay from A₀ to A₀ / 6 .

a.) To solve this, we have the following differential equation:

dA/dt = kA

With the initial condition A(0) = A₀

Rewriting the differential equation like this:

dA/A = kdt

And if we integrate both sides we get:

ln |A| = kt + c₁

Where  is a constant. If we apply exponential for both sides we get:

$A=e^{kt}e^{c} = C e^{kt}$

Using the initial condition A(0) = A₀ we get ,

A₀ = C

So the solution for the differential equation is given by:

$A(t) = A_{0}e^{kt}$

For the half life we know that we need to find the value of t for where we have,

A(t) = 1/2 (A₀)

Using this we get ,

$\frac{1}{2}A_{0} = A_{0}e^{kt$

Cancel A₀ on both sides and applying log on both sides we get ,

ln(1/2) = kt

t = {ln(1/2) / k } ----(1)

And using the fact that  ln(1/2) = -ln(2)  we get,

t = - { ln(2) / k }

b.) To solve this we consider ,

$A(t) = A_{0} e^{kt}$

Replacing k with value obtained from 1 we get,

k = - {ln(2) / T}

$A(t) = A_{0}e^{\frac{ln(2)}{T}t$

Cancel the exponential with the natural log, we get,

$A(t) = A_{0} 2^{-\frac{1}{T} }$

c.)For this case we find the value of t when we have remaining A₀/6

So we can use the following equation:

A₀/6 =$A_{0}2^{-\frac{t}{T} }$

Simplifying we got:

1/6 = $2^{-\frac{t}{T} }$

We can apply natural log on both sides and we got:

ln (1/6) = -t/T{ln(2)}

And if we solve for t we got:

t = T { ln(6) / ln(2) }

We can rewrite this expression like this:

t = T { ln(2²°⁵) / ln(2) }

Using properties of natural logs we got:

t = 2.5T { ln(2) / ln(2) }

t = 2.5T

Thus it will take 2.5 T for the substance to decay from A₀ to A₀/6 

Solve more problems on Half-life at :brainly.com/question/16439717

#SPJ4