P3 Differential Eqn (Boo)

A lump of radioactive substance is disintegrating. At time t days after it was first observed to have mass 10 grams, it's mass is m grams AND;

dm/dt = -km where k is a positive constant.

Find the time in days for the substance to reduce to 1 gram in mass. Given that the half-life is 8 days.

I think I've worked out that m = Ae^-kt

Not sure where to go next :confused:

Any help would be appreciated :smile:

Reply 1

Newton

Let "log(n)" denote the natural logarithm of n.

(dm/dt)=-km

=>dm=-km dt

=>INT(1/m)dm=INT-k dt

=>log|m|=-kt+C

Raising each side to e:

m=(e^(-kt+C))=(e(-kt))(e^C)

When t=0:

m=(e^(-0+C))=>m=(e^C)

Because this condition is for t=0 when there is an initial amount of substance, m_ it implies that the set constant is equal to the initial amount of substance:

=>m=m_(e^(-kt))

They say that the half life is 8 days, this means that there shall be one half of the substance left when t=8, therefore m shall only be a half of the initial amount of the substance:

=>(1/2)=(e^(-8k))

=>log(1/2)=-8k

=>-log2=-8k

=>log2=8k

=>k=(log2/8)~0.087

Hence the initial equation is now:

m=m_(e^(-0.087t))

The initial mass is 10g

=>m=10(e^(-0.087t))

They ask you for how long is shall take for the mass to equal to 1g:

=>1=10(e^(-0.087t))

=>(1/10)=(e^(-0.087t))

Taking logartihms of each side:

=>log(1/10)=-0.087t

=>-log10=-0.087t

=>log10=0.087t

=>((log10)/(0.087))=t

=>26.5=t

=>t=26.5days~27days

Newton.