Exponential decay help

sam2109

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(edited 13 years ago)

Reply 1

charco

Study Forum Helper

Original post by sam2109

hi people im not sure if im doing these right or wrong, inputted will be appreciated:
The half life of caseium is 42 years,
calculate the K for Caseium
i got Ln0.5 / 42 = 0.0165? is it right

the next questionalculate the percentage of the original caseium that remains after 100 years?

i got 5.21

next question is calculate the time it takes for the amount of casium to fall 5% of its original value.
i done ln0.05 / 42 - 181.56 hrs

can anyone correct me please

t_1/2 = 0.693/k

k = 0.693/42 = 0.0165 :smile:

N_t = N₀e^-kt

So, when t = 100

proportion remaining = e^-kt = 0.192

100 * N_t/N₀ = 0.192 * 100 = 19.2%

checksum -------------------------------------------------------------------

One half life decreases count by 50%

42 year half life in 100 years makes 2 and a bit half lives = between 12.5 and 25% - therefore reasonable answer.

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If it falls to 5% then N_t/N₀ = 0.05

e^-kt = 0.05

-kt = -3

therefore t = 3/0.0165 = 181.6 years

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5% falls between four and five half lives (50%, 25%, 12.5%, 6.25%, 3.125%)

four half lives = 4 x 42 = 168 years
five half lives = 5 x 42 = 210 years

Therefore 181.6 years is reasonable :smile:

Reply 2

sam2109

........

(edited 13 years ago)

Reply 3

charco

Study Forum Helper

Original post by sam2109

thanks very much for helping identify my answer.
i would like to know how u got 19.2%?, show me the procedure, once again thnaks for ur help

I've shown you the working!

You start with the decay equation:

N_t = N₀e^-kt

This says that the activity at time = t is equal to the activity at time = 0 (the initial activity) multiplied by the term e^-kt

You simply rearrange this and substitute in the values for t and k.

You are told that time t = 100

You have already worked out k = 0.0165

You assume that N₀ is 100% activity (it's the initial value)

So you can work out N_t

Reply 4

sam2109

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(edited 13 years ago)

Reply 5

charco

Study Forum Helper

Original post by sam2109

i get a different answer
0.693*2.718^-0.0165*100 =13.3???????????

I don't know where you get this from! Why have you multiplied by 0.693 (ln2) - it doesn't appear in the equation.

If you work out 'kt' first you get 0.0165 x 100 = 1.65

Therefore e^-1.65 = N_t/N₀

0.192 = N_t/N₀

So if N₀ is taken as 100% then N_t = 19.2%

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Do you understand my checksum?

If you do one for your answer (13.3%)

Half lives give a percentage remaining according to the number of half-lives passed:

1 half-life = 50%
2 half-lives = 25%
3 half-lives = 12.5%

Your value of 13.3 is very close to 12.5 and so you would expect just less than three half lives to have passed. But 3 x 42 = 126, which is much larger than 100 (more than a half half-life bigger). It makes your answer very improbable (wrong)

It is always worthwhile looking for a way of estimating the probability of an answer being reasonable (checksum) to give confidence in your answer and to make sure you haven't made a mistake in your assumptions or calculations.