Question

the rate of decay of a radioactive substance is proportional to the remaining amount of the substance. write differential equation to describe the phenomenon, using K as the constant of proportion. the half lift T of the radioactive substance is defined as the time that elapse before the substance decays to half its amount. show that
K=(1/T)*ln2
if the initial level of radioactive substance of ra radioactive substance is 1 curie and its half life is 10 days, determine
(i) the percentage of the radioactive substance that remains after 100 days.

Answer 1

(ii) minimum storage period of the radioactive substance if it can only be disposed of when its level of radioactivity has reduced to less then 10^-4 curie

Answer 2

to start you off, let N be the number of radioactive atoms,

dN/ dt = - k N

dN/N = - k dt

can integrate both sides easily.

Answer 3

okay then can u completed plz

Answer 4

It is being reduced by half every ten days, and 100 days is ten of these periods, so it will
radioactivity = (1 curie) (0.5)^10 = 0.000977 curies.

Answer 5

ii) minimum storage period of the radioactive substance if it can only be disposed of when its level of radioactivity has reduced to less then 10^-4 curie

Answer 6

what about this

Answer 7

After you've integrated both sides to solve the differential equation, you'll have a formula for the radioactivity of the substance at time \(t\). Set that equal to \(10^{-4} \text{ curie}\) and solve for \(t\). That's the amount of time it will take to get to that level.

Answer 8

(i) the percentage of the radioactive substance that remains after 100 days.
okay for this one what i have to do i cant get it

Answer 9

Douglas computed the amount of radioactivity for you already. You just need to express it as a percentage of the original amount.

Answer 10

can u do it for me coz i cant get just the first one the second i will try to do it

Answer 11

Can I do what for you? Please use the proper spelling of words if you wish to communicate with me. There's no 140-character limit here.

Answer 12

i mean after i have integrated both sides can u apply the conditions and do it for me
i mean this one
(i) the percentage of the radioactive substance that remains after 100 days.

Answer 13

As I said, Douglas already did this:

"It is being reduced by half every ten days, and 100 days is ten of these periods, so it will
radioactivity = (1 curie) (0.5)^10 = 0.000977 curies."

However, if you want it the other way:
\[\frac{dN}{N} = - k dt\]\[\int \frac{dN}{N} = -k \int dt\]\[\ln N = -kt + C\]Solving that for \(N\) we have
\[N = Ce^{-kt}\]At \(t=0\)
\[1 = Ce^{-k0}\]\[C=1\]
We find \(k\) with the formula provided:
\[k = \frac{1}{T} \ln 2 = \frac{1}{10}\ln 2 \approx 0.0693147\]So our general formula for the amount of radioactivity at time \(t\) is
\[N(t) = e^{-0.0693147t}\]

If we try it out with \(t=100\) to check Douglas' work:
\[N(100) = e^{-0.0693147*100} \approx 0.000976564\]

Now, you use that formula to find the value of \(t\) such that \[N(t) = 10^{-4}= e^{-0.0693147t}\]