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Question

Find the half life of a radioactive element, if its activity decreases for 1 month by 10%.

A
193.3 days
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B
197.3 days
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C
198.5 days
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D
199.7 days
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Solution

The correct option is C 197.3 days

Activity of an isotope is measured by the number of nuclei decaying for a time unit.

Suppose that dNd nuclei decay for a short period of time dt.

Then the isotope activity A is expressed by the formula A=dNddt

It follows from the radioactive decay law that

N(t)=N0eλt where N(t) is the quantity of the remaining nuclei.

Therefore,

Nd(t)=N0N(t)=N0N0eλt=N0(1eλt)

By differentiating with respect to t, we find the expression for activity:

A(t)=dNddt=N0λeλt

The initial isotope activity is equal to

A(t=0)=A0=N0λ

Hence, A(t)=A0eλt

As it can be seen, the activity decreases over time by the same law as the amount of undecayed material. Substituting the expression for the half life T=ln2λ in the last formula, we can write:

A(t)=A0etln2T

The value of T in the last expression can be found by

etln2T=AA0

tln2T=lnAA0

tln2T=lnA0A

T=tln2lnA0A

In our case, the half life period of the given isotope is

T=tln2lnA0A=30ln2ln10090300.93ln1.11197.3days


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