Question

The half-life of a radioactive substance against a-decay is \[1.2 \times {10^7}\,{\mathop{\rm s}\nolimits} \]. What is the decay rate for \[4 \times {10^{15}}\] atoms of the substance?
A. \[4.6 \times {10^{12}}\,atoms/{\mathop{\rm s}\nolimits} \]
B. \[2.3 \times {10^{11}}\,atoms/{\mathop{\rm s}\nolimits} \]
C. \[4.6 \times {10^{10}}\,atoms/{\mathop{\rm s}\nolimits} \]
D. \[2.3 \times {10^8}\,\,atoms/{\mathop{\rm s}\nolimits} \]

Answer 1

Hint: According to Rutherford and Soddy, the law of radioactive decay says “ at any instant the rate of decay of a radioactive atom is proportional to the number of atoms present at that instant”. Here, we have to apply the equation of disintegration to find the final answer.

Formula used:
\[ - \dfrac{{dN}}{{dt}} \propto N \Rightarrow \dfrac{{dN}}{{dt}} = - \lambda N\] and \[N = {N_0}{e^{ - \lambda t}}\]
Here, N = Number of atoms at time t, \[{N_0} = \] Initial number of atoms present and \[\lambda = \,\] Decay Constant.

Complete step by step solution:
As the number of atoms N in a radioactive element present at any instant t is,
\[N = {N_0}{e^{ - \lambda t}}\,......(1)\]
Differentiating the above equation with time we get,
Decay rate, \[\dfrac{{dN}}{{dt}} = {N_0}{e^{ - \lambda t}}\]
\[ \Rightarrow \dfrac{{dN}}{{dt}} = - \lambda {N_0}{e^{ - \lambda t}} \\ \]
\[\Rightarrow \dfrac{{dN}}{{dt}} = - \lambda N\,\,.....(2)\].............(As, \[N = {N_0}{e^{ - \lambda t}}\])
Decay constant has value where T is half life.
Substituting \[\lambda = \dfrac{{0.693}}{T}\] in equation (2) we get,
\[\dfrac{{dN}}{{dt}} = \dfrac{{0.693}}{T} \times N\,.....(3)\]

Given here, we have N = \[4 \times {10^{15}}\] and T =\[1.2 \times {10^7}\,{\mathop{\rm s}\nolimits} \].
Substituting values of N and T in equation (3) we get,
\[\dfrac{{dN}}{{dt}} = \dfrac{{0.693}}{{1.2 \times {{10}^7}\,}} \times 4 \times {10^{15}}\,\\
\therefore \dfrac{{dN}}{{dt}} = 2.3 \times {10^8}\,atoms/\sec \]
Hence, the decay rate for \[4 \times {10^{15}}\,\] atoms of the given substance will be \[2.3 \times {10^8}\,atoms/\sec \].

Hence, option D is the correct answer.

Note: Radioactive decay is a spontaneous process which means that it is impossible to predict when an individual atom will decay and half life of radioactive substances can be changed by using time dilation effect which has been verified many times in particle accelerators.