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Question: A radioactive substance has \[{10^8}\] nuclei. Its half-life is \[30\,{\text{s}}\]. The number of nu...

A radioactive substance has 108{10^8} nuclei. Its half-life is 30s30\,{\text{s}}. The number of nuclei left after 15s15\,{\text{s}} is nearly
A. 2×1052 \times {10^5}
B. 3×1063 \times {10^6}
C. 7×1077 \times {10^7}
D. 5×1085 \times {10^8}

Explanation

Solution

Use the formula for the population of the radioactive substance at any time t. This formula gives the relation between the population of the radioactive substance at any time, initial population of the radioactive substance, time and half life period of the radioactive substance. Substitute all the values in this formula and calculate the required answer.

Formula used:
The population of the radioactive substance NN at any time is given by
N=N0(12)tTN = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}} …… (1)
Here, N0{N_0} is the initial population of the radioactive substance, tt is the time and TT is the half-life period of the radioactive substance.

Complete step by step answer:
We have given that the initial population of the radioactive nuclei is 108{10^8}.
N0=108{N_0} = {10^8}
The half-life period of the radioactive substance is 30s30\,{\text{s}}.
T=30sT = 30\,{\text{s}}
We have asked to determine the population of the radioactive nuclei at time 15s15\,{\text{s}}.
t=15st = 15\,{\text{s}}
Substitute 108{10^8} for N0{N_0}, 15s15\,{\text{s}} for tt and 30s30\,{\text{s}} for TT in equation (1).
N=(108)(12)15s30sN = \left( {{{10}^8}} \right){\left( {\dfrac{1}{2}} \right)^{\dfrac{{15\,{\text{s}}}}{{30\,{\text{s}}}}}}
N=(108)(12)12\Rightarrow N = \left( {{{10}^8}} \right){\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{2}}}
N=(108)12\Rightarrow N = \left( {{{10}^8}} \right)\sqrt {\dfrac{1}{2}}
N=1082\Rightarrow N = \dfrac{{{{10}^8}}}{{\sqrt 2 }}
N=1081.414\Rightarrow N = \dfrac{{{{10}^8}}}{{1.414}}
N=0.70×108\Rightarrow N = 0.70 \times {10^8}
N=7×107\therefore N = 7 \times {10^7}

Therefore, the number of radioactive nuclei left after 15 seconds is nearly 7×1077 \times {10^7}.Hence, the correct option is C.

Additional information:
The process in which the nucleus of a radioactive atom undergoes disintegration spontaneously is known as radioactive decay.The time required for half of the initial number of nuclei of the radioactive substance to decay is known as the half-life period of the radioactive substance.

Note: One can solve the same question by another method. One can determine the value of the decay constant from the half-life period of the radioactive substance and then use the decay rate formula to determine the number of nuclei left at any time after the disintegration process has started.