Question

The half life of the radioactive material radium-226 is 1590 years.
(a) A sample of the radium-226 has a mass of 100mg. Find a formula for the mass of radium-226 that remains after t years.

(b) FInd the mass after 1000 years correct to the nearest milligram.

(c) when will the mass be reduced to 30 mg?

(d) At what rate is the mass decreasing when 30 mg remains?

Answer 1

\[\large A=A _{0}*2^{-t/h}\]

A=final amount
A_o=initial amount
t=time
h=half life

wanna plug some things in that we know?

Answer 2

But isn't the formula for decay \[A(t)= C e ^{kt}\] ?

Answer 3

So then when i plug things in it would be \[100e ^{kt}\] but i dont know what to do after.

Answer 4

That does look more familiar. Im probably mixing things up, well go with yours.

Answer 5

One second while I refresh myself I remember doing this last semester

Answer 6

Okay thanks!

Answer 7

ok i remember.



we know an obvious solution, since the half life is 1590 years and we start with 100mg, after 1590 years we will have 50mg. so basically that gives us some variables we can plug in then solve for k. once we know k well have the formula and can solve for the amount at any given time

50=100e^(1590k)

so solve for k in that equation

Answer 8

so would it be \[\ln \frac{ 1 }{ 5 }\ /1590\] ?

Answer 9

i mean 1/2

Answer 10

1/2=e^1590k

ln1/2=1590k

(ln1/2)/1590=k

that is correct. so now that we know that, we can set our equation up

\[A(t)=100e^{((\ln1/2)/1590))t}\]

so theres the answer to a

Answer 11

For b, you just wanna plug in 1000 for t, and solve

Answer 12

Would the answer be? \[a(t)= 1000e ^{\ln \frac{ 1 }{ 2 }/1590 (1000) t}\]\[= 64.7 mg\]

Answer 13

and sorry i mean 100e

Answer 14

yep so to the nearest milligram, 65mg

c) plug 30 in for the amount and solve for t

Answer 15

I know I have to take the logarithm of both sides so one side would be ln 30 but how would i take ln from \[100e ^{\frac{ \ln 1/2 }{ 1590}(t)}\]

Answer 16

give me a second, doing it on paper

Answer 17

Thanks

Answer 18

ok, what you do is this.

30=100e^((ln1/2)/1590)t

divide by 100

3/10=e^((ln1/2)/1590)t


take ln of both sides, remember ln is natural log and the base is 'e', so the ln and e cancel out

ln 3/10=((ln 1/2)/1590)t


divide both sides by ((ln(1/2)/1590), when you divide by a fraction you multiply by the reciprocal

(ln 3/10)*((1590/ln(1/2))=t

wanna finish?

Answer 19

so t = (1590)(ln 3/10) / (ln 1/2) or can it be simplified even further?

Answer 20

I think thats as good as it gets. slap that in a calculator to get your number answer

Answer 21

it would be around 2762 years

Answer 22

Correct. I'm not sure about d. Do you have any ideas?

Answer 23

No, I have no idea. Oh well. You've helped me enough. Thanks for your help!

Answer 24

You're welcome