Question

A radioactive isotope has a half-life of T years. How long will it takethe activity to reduce to a) 3.125%, b) 1% of its original value?

Answer 1

a)

Given: The half life of the radioactive isotope is T years and the reduction in the activity is 3.125%.

According to the law of radioactive decay,

N= N 0 e −λt N N 0 = e −λt

Where, N is the remaining number of molecules, N 0 is the initial number of atoms, λ is the decay constant and t is the time.

By substituting the given values in the above expression, we get

3.125 100 = e −λt 1 32 = e −λt λt= log e ( 32 ) t= 3.4657 λ

We know that,

λ= 0.693 T

By substituting the given value in the above expression, we get

t= 3.466 0.693 T =5T years

Thus, the isotope will take about 5T years to reduce to 3.125% of its original value.

b)

Given: The reduction in the activity is 1%.

According to the law of radioactive decay,

N= N 0 e −λt N N 0 = e −λt

By substituting the given values in the above expression, we get

1 100 = e −λt 1 100 = e −λt λt= log e ( 100 ) t= 4.6052 λ

We know that,

λ= 0.693 T

By substituting the given value in the above expression, we get

t= 4.0652 0.693 T =6.65T years

Thus, the isotope will take about 6.65T years to reduce to 1% of its original value.