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Question: Calculate the half-life period of a radioactive substance if its activity drops to \(\dfrac{1}{{16}}...

Calculate the half-life period of a radioactive substance if its activity drops to 116\dfrac{1}{{16}}th of its initial value in 3030 years.

Explanation

Solution

By using the initial and the total activity of the substance we will find the total number of half lives by converting those values in the power of 12\dfrac{1}{2}. After that we will divide the total time with respect to the number of half lives and eventually find the half-life.

Complete step-by-step solution:
In nuclear physics, half-life is defined as the time required for a quantity of a substance to reduce to half of its initial quantity.
Activity of a substance is defined as the value which signifies the amount of any substance being decayed in unit time.
For the given set of problem we have to use the relation,
NN0=(12)n(1)\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^n} - - - - - (1)
The given variables here are defined as-
N=N = the final amount undecayed quantity.
N0={N_0} = the initial amount of substance.
n=n = number of half-life.
According to the given question, N=N016N = \dfrac{{{N_0}}}{{16}}
Substituting the value in equation (1)\left( 1 \right) we get,
(116)=(12)n\left( {\dfrac{1}{{16}}} \right) = {\left( {\dfrac{1}{2}} \right)^n}
Converting the value into the power of 12\dfrac{1}{2} we get,
(12)4=(12)n{\left( {\dfrac{1}{2}} \right)^4} = {\left( {\dfrac{1}{2}} \right)^n}
Comparing the equations we get, n=4(2)n = 4 - - - - \left( 2 \right).
So, we have got the number of half-life as 44 and thus using the total time as given in the question we will find the time period of half-life.
Total time period given in the question=30 = 30 years.
The relation is given as,
T12=tn(3){T_{\dfrac{1}{2}}} = \dfrac{t}{n} - - - - - - \left( 3 \right)
The variables are defined as-
T12{T_{\dfrac{1}{2}}} == half-life.
t=t = total time.
n=n = number of half-life.
Substituting the value of t=30t = 30 is given in the question and nnfrom (2)\left( 2 \right) into equation (3)\left( 3 \right) we get,
T12=304=7.5{T_{\dfrac{1}{2}}} = \dfrac{{30}}{4} = 7.5
Therefore, the half-life of this radioactive substance is 7.57.5 years.

Note: This method is only applicable when the ratio of initial activity to the final activity of a radioactive substance can be converted to the power of 12\dfrac{1}{2}. Either we have to use the Radioactive Decay Law to find the decay constant and hence find the half life.