Question

A radioactive material decays by simultaneous emission of two particles with the half lives 1620 yr and 810 yr respectively. The time in years after which one fourth of the material remains is.
a) 1080 yrs
b) 2340 yrs
c) 4860 yrs
d) 3240 yrs

Answer 1

Hint: In the question it is given that the radioactive element undergoes decay by simultaneous emission of two particles. Hence for such a situation the effective decay constant is given as the sum of the individual decay constant of the two particles. The half lives of the two particles is given and from that we can individual decay constant. Further we can use the radioactive decay law to determine the time for which only one fourth of the substance remains from the effective decay constant.

Complete step-by-step answer:
To begin with let us first define the half life of a radioactive element.
The half life of a radioactive element is defined as the time taken for that element to reduce to half of its initial composition. The half life of a radioactive particle is given by, ${{T}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{\lambda }$ where $\lambda $is the decay constant. Let us say the substance M with decay constant ${{\lambda }_{M}}$ emits the particle X whose half life is 1620 years and the other particle as Y whose half life is 810 yrs.
Hence from the above equation the decay constant of X i.e. ${{\lambda }_{X}}$ and the decay constant of Y i.e. ${{\lambda }_{Y}}$ is given by,
$\begin{align}
  & {{T}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{{{\lambda }_{X}}} \\
 & {{\lambda }_{X}}=\dfrac{0.693}{1620}=4.27\times {{10}^{-4}}\text{is the decay constant of X} \\
\end{align}$
Similarly decay constant of Y is,
$\begin{align}
  & {{T}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{{{\lambda }_{Y}}} \\
 & {{\lambda }_{Y}}=\dfrac{0.693}{810}=8.55\times {{10}^{-4}}\text{is the decay constant of Y} \\
\end{align}$
Hence the effective decay constant of the above substance is given by,
${{\lambda }_{EFFECTIVE}}={{\lambda }_{X}}+{{\lambda }_{Y}}$
Substituting the above decay constant of X and Y we get,
$\begin{align}
  & {{\lambda }_{M}}={{\lambda }_{EFFECTIVE}}={{\lambda }_{X}}+{{\lambda }_{Y}} \\
 & {{\lambda }_{M}}=4.277\times {{10}^{-4}}+8.555\times {{10}^{-4}}=12.832\times {{10}^{-4}} \\
\end{align}$
Now we are ready to calculate the time for which only one fourth of the substance M remains. The radioactive decay law is given by,
$\text{N=}{{\text{N}}_{\text{o}}}{{e}^{-\lambda t}}$ where N is the amount of substance remaining at time t, ${{\text{N}}_{\text{o}}}$is the amount of substance present at beginning of the decay and $\lambda $ is the decay constant.
It is asked in the question how long it will take such that only one fourth of the substance remains as that of the beginning. Using radioactive decay law we get,
$\begin{align}
  & \text{N=}{{\text{N}}_{\text{o}}}{{e}^{-\lambda t}} \\
 & \Rightarrow \dfrac{{{\text{N}}_{\text{o}}}}{4}={{\text{N}}_{\text{o}}}{{e}^{-12.82\times {{10}^{-4}}t}} \\
 & \Rightarrow \dfrac{1}{4}={{e}^{-12.82\times {{10}^{-4}}t}} \\
 & \Rightarrow {{e}^{12.82\times {{10}^{-4}}t}}=4 \\
 & \Rightarrow 12.832\times {{10}^{-4}}t=\ln (4) \\
 & \Rightarrow t=\dfrac{1.3863}{12.832\times {{10}^{-4}}}=1080yrs \\
\end{align}$
Hence from this we can conclude that the correct answer is option a.

So, the correct answer is “Option A”.

Note: It is to be noted that the 3 digits have to be taken after the decimal in any number while solving problems on decay of a substance. This is because rounding up may result in slight change in the answer and the answer one may get wrong although your method is correct. After taking three digits after the decimal, if the resultant answer comes in the form of decimal then ignore the digits after the decimal point while calculating the number of years.