Question

If \[{C^{14}}\] has half-life 5700 years. At 11400 years actual amount left is
A. 0.5 of original amount.
B. 0.25 of original amount.
C. 0.125 of original amount.
D. 0.625 of original amount.

Answer 1

Hint: Half-life of a radioactive substance is defined as the time interval in which its mass reduces to half of the initial mass and at any instant rate of decay of radioactive atoms is directly proportional to the number of atoms present at that state.

Formula used:
\[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^n}\]
Where, N = Number of atoms after time t, \[{N_0} = \] Number of atoms at time t= 0 and n = Ratio of given time with half-life.

Complete step by step solution:
Given here is \[{C^{14}}\] with a half-life of 5700 years we have to find the number of atoms present after 11400 years.

As we know the ratio of the number of atoms (N) after time t with the initial number of atoms \[\left( {{N_0}} \right)\] can be given by the relation,
\[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^n}\,.......(1)\]
Where, N = Number of atoms after time t, Number of atoms at time t= 0 and n = Ratio of given time with half-life.

Before substituting the given values in equation (1) we need to calculate n, which is equal to
\[n=\dfrac {t}{t_{\frac {1}{2}}}=\dfrac {11400}{5700}=2\]

Now, substituting values in equation (1) we get,
\[\begin{array}{l}\dfrac{N}{{{N_0}}} = \,{\left( {\dfrac{1}{2}} \right)^2}\, \Rightarrow \,\dfrac{N}{{{N_0}}}\, = \,\dfrac{1}{4}\\ \Rightarrow \,N\, = \,0.25\,{N_0}\end{array}\]

From the above relation, it is clear that the actual amount of \[{C^{14}}\]left after 11400 years is 0.25 of its original amount.

Therefore, option B is the correct option.

Note: If a radioactive substance has only one atom then it is not accurate to say that only half the atom will remain after half-life, the half-life of an atom gives the average time for the atom to decay in its half or we can say that probability of that substance to decay within half-life is 50%.