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I know this problem has to do with half-life but I do not know how to do this problem. Thanks for any help. It is gratly appreciated.

a) Find the activity of a sample of 3 x 1016 nuclei of 226Ra which has a half-life of 1600 years.

b) How long will it take for the number of nuclei to be reduced by 10%?

c) What will be the activity at this time?

The decay constant lambda of a radioactive isotope is equal to 0.693 divided by the half lifeT1/2:

lambda=0.693/T1/2=0.693/(1600yr*365days/yr*24hrs/day*3600sec/hr)=1.37x10^-11/sec

[The decay constant is the probability that any particular radioactive atom will decay in the given time period.

The activity Ao of a radioactive sample is equal to the product of the number of atoms present No multiplied by the decay constant lambda.

Ao=No*lambda=3.0x10^16*1.37x10^-11/sec=4.12x10^5/sec=4.12x10^5Bq

The activity A of a sample at any future time t is equal to:

A=Ao*e^(-lambda*t)

In this case the new activity A is 90% of the initial activity Ao:

0.90Ao=Ao*e^(-lambda*t)

0.90=e^(-lambda*t)

Take the natural log of both sides of,the equation:

ln[0.90=e^(-lambda*t]

-0.105=-lambda*t=1.37x10^-11*t

Therefore, the time t when the activity A, and therefore the number of atoms present is 90% of the number initially present becomes:

t=7.66**10^9sec=243yrs

While the activity will be 90% of the initial activity:

0.90*4.12x10^5=3.71x10^5Bq

Physics

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