Question

A radioactive material decays at a rate proportional to the amount present.
Initially there are 100 milligrams of the material present and after 1 hour(s)
the material has 20 milligrams.


Find the material decay constant,

Answer 1

You need to solve:\[\frac{ dm }{ dt }=k·m \rightarrow \frac{ dm }{ m }=k·dt \rightarrow Ln[m(t)]=k·t+A \rightarrow m(t)=e^{A}e^{k·t}=Ce^{k·t}\]\[m(0)=100=C \rightarrow m(t)=100e^{k·t}\]\[m(1)=100e^{k}=20\rightarrow e^k=\frac{ 1 }{ 5 }\rightarrow k=-Ln(5)=-1.609\]Units of "k" are in \[[hours]^{-1}\]

Answer 2

i didnt mean to close the question

Answer 3

i had 1.609 but it told me i was wrong

Answer 4

It has a negative sign.

Answer 5

i tried both

Answer 6

then their answer is wrong or you are missing something (do you have to put the units?)

Answer 7

no i dont need units

Answer 8

Just try:\[m(t)=100e^{-1.609·t}\]at t=0 and t=1 and check it. Also\[\frac{ dm(t) }{ dt}=-1.609·100·e^{-1.609·t}=-1.609·m(t)\]All conditions are fulfilled with our solution

Answer 9

Hoow many decimal places you have to put? If it is 2, just put -1.61.If it is one, put -1.6

Answer 10

is that the mass of the material as a function of time?

Answer 11

of course. THe problem states that mass decays proportionally to the amount present and that amount has changed in time

Answer 12

Find the mass of the material N_1 after 5 hours,