Question

Plutonium decays with half lifetime 24000 yrs. If Plutonium is stored after 72000 years, the fraction that remains is,
A. 1/2
B. 1/4
C. 1/6
D. 1/8

Answer 1

We all know that the half-life has a probabilistic nature and denotes the time in which, on average, about half of entities decays. If we had only one atom, then it is not like after one half-life, one-half of the particles will decay, so we can say that half-life describes the decay of distinct entities.

Complete step by step answer:
The half-life of a radioactive substance can be defined as the time taken for a given amount of the substance to become reduced by half due to decay, and therefore, the emission of radiation.
We know that the rate of disintegration of a radioactive particle is expressed as:
$N = {N_0}{e^{ - \lambda t}}$
Here N is the number of particles at a given time t and ${N_0}$ is the initial number of particles, $\lambda$ is the decay constant, and t is the given time.
We will now express the formula for the disintegration of the particles in terms of its half-life.
$N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}\,......\,{\rm{(I)}}$
Here, T is the half-life.
We will now substitute t=72000 yrs and T=24000 yrs in equation (I) to find the fraction left value.

\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{72000\;{\rm{yr}}}}{{24000\;{\rm{yr}}}}}}\\\ = {\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{3}}}\\\ = \dfrac{1}{8} \end{array}$$ Therefore the fraction of Plutonium left is $$\dfrac{1}{8}$$, and the correct option is (D). **Note:** Decay constant is the proportionality constant between the size of a number of radioactive atoms and the rate at which the number of atoms decreases because of radioactive decay. Throughout the reactive disintegration, the disintegration constant $\lambda $ remains constant. Any radioactive decay follows first-order kinetics and is related to the half-life as given in the formula below. That $\lambda $ then can be used to find the time, and the relation to time is also shown below.