Question

A freshly prepared sample of a radioisotope of half-life 1386s has activity ${{10}^{3}}dps$ Given that $\ln 2=0.693$, the fraction of initial number of nuclei (expressed in nearest integer percentage) that will decay in first 80s after preparation of the sample is ___.

Answer 1

Here we have given a radioactive sample whose fraction of initial number of nuclei which will decay in the 80s is to be found. By using the formula for the number of nuclei undecayed we can find the number of nuclei decayed after the 80s and so the fraction. We have given all the parameters which are required to solve the question.
Formula used:
$N={{N}_{0}}{{e}^{-\lambda t}}$

Complete answer:
A radioactive substance is the one which has unstable nuclei and to become stable it decays until it gains the stability. The number of nuclei undecayed at time t or the number of nuclei left at time t is given as
$N={{N}_{0}}{{e}^{-\lambda t}}$
Where ${{N}_{0}}$is the initial number of nuclei or the amount of nuclei present at time $t=0$and λ is the decay constant.
Now the amount of nuclei decayed can be given as ${{N}_{0}}-N$. So the fraction asked here can be given as

& \text{Fraction}=\dfrac{{{N}_{0}}-N}{{{N}_{0}}} \\\ & \Rightarrow \text{Fraction}=\left( 1-\dfrac{N}{{{N}_{0}}} \right) \\\ \end{aligned}$$ Now substituting value of N in above equation we get $$\begin{aligned} & \text{Fraction}=\left( 1-\dfrac{{{N}_{0}}{{e}^{-\lambda t}}}{{{N}_{0}}} \right) \\\ & \Rightarrow \text{Fraction}=\left( 1-{{e}^{-\lambda t}} \right)\text{ }........\text{(i)} \\\ \end{aligned}$$ Now the decay constant λ is given by the following formula $$\lambda =\dfrac{0.623}{{{t}_{\dfrac{1}{2}}}}$$ Where $${{t}_{\dfrac{1}{2}}}$$ is the half-life of the substance. For the given substance half-life is $${{t}_{\dfrac{1}{2}}}=1386s$$, so the value of decay constant can be given as $$\begin{aligned} & \lambda =\dfrac{\ln 2}{1386} \\\ & \Rightarrow \lambda =\dfrac{0.623}{1386} \\\ & \Rightarrow \lambda =0.00045{{s}^{-1}} \\\ \end{aligned}$$ Substituting the decay constant in equation (i) and $$t=80s$$as we have to find fraction at 80s, we get $$\begin{aligned} & \text{Fraction}=\left( 1-{{e}^{-\left( 0.00045 \right)\left( 80 \right)}} \right) \\\ & \Rightarrow \text{Fraction}=1-{{e}^{-0.036}} \\\ & \Rightarrow \text{Fraction}=1-0.96 \\\ & \Rightarrow \text{Fraction}=0.04 \\\ \end{aligned}$$ Now it is asked to express it in the nearest integer percentage, hence the fraction is 4%. **Note:** We can also conclude that in the 80s, 4% of the substance decayed. While calculating decay constant we used the value of ln 2 which was given in question. Note that in the radioactive process $\lambda$ is decay constant and not the wavelength of the particle. We were also given an activity through which we could have found the ratio if the value of $${{N}_{0}}$$was given.