Question

Calculate the time required to decay a $\dfrac{15}{16}$ of the radioactive element if half life of the element is 1600 years.
A) 1000 years
B) 1500 years
C) 1600 years
D) 3200 years
E) 6400 years

Answer 1

Hint: Law of radioactive decay is used to solve this question. Radioactive reactions are first order reactions. Half life time of radioactive is given which can be used to find the total time required to decay the given radioactive element via radioactive decay constant (lambda).

Formula used:
${{\text{t}}_{\dfrac{\text{1}}{\text{2}}}}\text{=}\dfrac{\text{0}\text{.693}}{\text{ }\!\!\lambda\!\!\text{ }}$
$\text{ }\!\!\lambda\!\!\text{ =}\dfrac{\text{1}}{\text{t}}\text{ln}\dfrac{{{\text{N}}_{\text{0}}}}{\text{N}}$

Complete step by step answer:
Now, in this question the half life time of the radioactive element is given.
So by the first order reaction we can find the value of radioactive constant,
$\Rightarrow \text{ }\!\!\lambda\!\!\text{ =}\dfrac{\text{0}\text{.693}}{{{\text{t}}_{\dfrac{\text{1}}{\text{2}}}}}$
${{\text{t}}_{\dfrac{\text{1}}{\text{2}}}}\text{=1600years}$ (Given)
Now, let’s apply the value of half life in first order reaction formula,
$\Rightarrow \text{ }\!\!\lambda\!\!\text{ =}\dfrac{\text{0}\text{.693}}{1600}$
$\Rightarrow \text{ }\!\!\lambda\!\!\text{ }=4.33\times {{10}^{-4}}$

As we know according to the law of radioactive decay, we have one formula,
$\text{N(t)=}{{\text{N}}_{\text{0}}}{{\text{e}}^{\text{- }\!\!\lambda\!\!\text{ t}}}$
Here,
\[\text{ }\!\!\lambda\!\!\text{ =}\] Radioactive decay constant or disintegration constant
t=time
${{\text{N}}_{\text{0}}}=$ Number of nuclei at time t
N = Number of nuclei in sample before decay
We can now simply the above formula as,
$\text{ }\!\!\lambda\!\!\text{ =}\dfrac{\text{1}}{\text{t}}\text{ln}\dfrac{{{\text{N}}_{\text{0}}}}{\text{N}}$
Now, to find the time, lets reframe the formula,
$\text{t=}\dfrac{\text{1}}{\text{ }\!\!\lambda\!\!\text{ }}\text{ln}\dfrac{{{\text{N}}_{\text{0}}}}{\text{N}}$
Here,
${{\text{N}}_{\text{0}}}$ We don’t know the exact value, so we will use it as it is.
And,
$\begin{align}
  & \text{N=}{{\text{N}}_{\text{0}}}\text{-}{{\text{N}}_{\text{0}}}\left( \dfrac{\text{15}}{\text{16}} \right) \\
 & \text{N=}{{\text{N}}_{\text{0}}}\left( \dfrac{\text{1}}{\text{16}} \right) \\
\end{align}$
Now, let’s put the values in the above formula to get the answer of our question,
\[\text{t=}\dfrac{\text{1}}{4.33\times {{10}^{-4}}}\text{ln}\dfrac{{{\text{N}}_{\text{0}}}}{{{\text{N}}_{\text{0}}}\left( \dfrac{1}{16} \right)}\]
Now, as we can see the ${{\text{N}}_{\text{0}}}$ is common, we will eliminate it.
Hence, we will get,
\[\text{t=}\dfrac{\text{1}}{4.33\times {{10}^{-4}}}\text{ln}\left( 16 \right)\]
Now, let’s solve it further,
\[\text{t=}\dfrac{\text{1}}{4.33\times {{10}^{-4}}}\times 2.77\]
\[\Rightarrow \text{t=0}\text{.64032}\times \text{1}{{\text{0}}^{4}}\]
\[\Rightarrow \text{t = 6403}\text{.2years}\]
Now, we got the answer of time that is 6403.2 years.
As we can see in the given options, the nearest answer is 6400 years; we will consider the option E as our correct and most suitable answer.
Hence, the correct answer is option E) 6400 years.

Note:
The radioactive reactions are always first order reactions. We will prefer to assume the value of the number of nuclei at time t, as a variable rather than assuming any particular numerical value, when it is not given in the question. Don’t forget to multiply the fraction with the number of nuclei at time t and subtract it from the number of nuclei at time t, while calculating the number of nuclei in sample before decay.