Question

Strontium 90 is a radioactive material that decays according to the function A(t) = A0e ^-0.0244t where A0 is the initial amount present and A is the amount present at time t (in years) Assume that a scientist has a sample of 500 grams of strontium 90.
What is the decay rate of strontium 90?

Answer 1

\[N(t)=N_0e^{-\lambda t}\]
Where \(N(t)\) is the amount of \(N\) at time \(t\) ,
\(\lambda\) is the decay constant for the radioactive source

\[A(t)=A_0e^{-0.2044t}\]\[A(0)=A_0e^{-0.2044\times 0}=A_0=500\text g\]

Answer 2

Is this the whole question?
seams like some bits are missing,

Answer 3

this is the who question

Answer 4

whole

Answer 5

so you can find the decay rate/

Answer 6

radioactive decay is a first order reaction so the decay is at a constant rate

Answer 7

isn't 500 the amount present?

Answer 8

I think you have to use natural logs

Answer 9

Perhaps you are looking fot the Activity of the radioactive strontium

Answer 10

i'm looking for the rate of decay

Answer 11

The activity is the number of decays per second ,so i guess you could call it the rate of decay

Answer 12

ok.....and how do you get the rate of decay

Answer 13

\[\text{rate of decay}=\text{Activity}=\lambda N(t)\]

Answer 14

that's not helping sorry

Answer 15

because
\[\text{rate of decay}=\text{Activity}=-\frac{ \text dN(t)}{\text dt}\]

Answer 16

well then how do you solve the problem. I'm not following

Answer 17

\[\text {activity}=-\frac{ \text dN(t)}{\text dt}=-\frac{ \text d}{\text dt}\left(N_0e^{-\lambda t}\right)\]

Answer 18

the decay rate means the rate of the number of decays ,
the negative of the amount of substance left after an amount of time

Answer 19

can you differentiate that ?

Answer 20

it's not a differential problem it's a natural log problem

Answer 21

\[-\frac{ \text d}{\text dt}\left(N_0e^{-\lambda t}\right)=\lambda e^{-\lambda t}=\lambda N(t)\]

Answer 22

\[N(t)=N_0e^{-\lambda t}\]\[\frac{N(t)}{N_0}=e^{-\lambda t}\]\[\ln\left(\frac{N(t)}{N_0}\right)=\ln e^{-\lambda t}\]\[\ln\left(\frac{N(t)}{N_0}\right)=-\lambda t\]\[\ln\left(\frac{N_0}{N(t)}\right)=\lambda t\]

Answer 23

i hope i havent confused you with so many equations

Answer 24

where you you put 500 grams in to the equation?

Answer 25

\(500 \text g\) is amount of strontium at \(t=0\)
ie \[N(0)=N_0=500 \text g\]

Answer 26

ok. and how do i solve the last step?

Answer 27

what are you up to

Answer 28

\[\text{activity}=\lambda N(t)=\lambda N_0e^{-\lambda t}\]\[=0.0244\left[\frac1{\text {yr}}\right]\times 500 [\text g]\times e^{-0.0244t}\]\[=\cdots\left[\frac{\text g}{\text{ yr}}\right]\]

Answer 29

the specific activity is
\[\lambda N_0\]

Answer 30

which would just be
\[=0.0244\left[\frac1{\text {yr}}\right]\times 500 [\text g]=\cdots \left[\frac {\text g}{\text {yr}}\right]\]

Answer 31

im not sure which the question is looking for.