Question

Polonium-214 has a relatively short half life of 164 s. How many seconds would it take for 8.0 g of this isotope to decay to 0.25 g?

Answer 1

In a radioactive decay, the half life is defined as the period of time after which there is a 50% chance that an atom will have undergone nuclear decay. At this point of time, the amount left is just exactly half of the initial amount.

Formula used: We would require the following formulas:
$\dfrac{{{N}_{o}}}{N}={{2}^{n}}$
$n=\dfrac{time}{{{T}_{1/2}}}$

Complete step by step answer: -The spontaneous breakdown of the nucleus of an atom of a radioactive substance resulting in the emission of radiation from the nucleus is known as Radioactive decay.
-Half-life (denoted as${{T}_{1/2}}$) is defined as the time required for a quantity to reduce to half of its initial value.

-We will use the following formula to calculate number of periods:-
$\dfrac{{{N}_{o}}}{N}={{2}^{n}}$
where,
${{N}_{o}}$ = the initial mass of Polonium-214
N = the mass of Polonium-214 left after decay in certain time period
n= number of periods
We have been provided the following values:-
${{N}_{o}}$ = the initial mass of Polonium-214 = 8.0g
N = the mass of Polonium-214 left after decay in certain time period = 0.25g
Therefore, ${{2}^{n}}=\dfrac{8.0g}{0.25g}$
$\begin{aligned} & \Rightarrow {{2}^{n}}=32 \\\ & \Rightarrow {{2}^{n}}={{2}^{5}} \\\ & \Rightarrow n=5 \\\ \end{aligned}$
-Calculation of time Polonium-214 took to decay to 0.25 g:-
$n=\dfrac{time}{{{T}_{1/2}}}$
where,
${{T}_{1/2}}$= half life of the substance = 164 seconds.
$\begin{aligned} & \Rightarrow n=\dfrac{time}{{{T}_{1/2}}} \\\ & \text{On rearranging it, we get: time}=n\times {{T}_{1/2}} \\\ & \Rightarrow time=5\times 164\operatorname{s} \\\ & \Rightarrow time=820s \\\ \end{aligned}$
-Hence, Polonium-141 will take 820 seconds for decaying of 8.0 g of this isotope to 0.25 g.

Note: -The alternative method to solve this question is shown below:-
In this method we will keep doing the half of the initial amount till we reach 0.25g, as we are using ${{T}_{1/2}}$time to divide the initial amount into halves. It is illustrated as follows:-
$8g\xrightarrow[1]{}4g\xrightarrow[2]{}2g\xrightarrow[3]{}1g\xrightarrow[4]{}0.5g\xrightarrow[5]{}0.25g$
Number of times the amount is divided into half (or number of time periods) = 5.
$\begin{aligned} & \Rightarrow n=\dfrac{time}{{{T}_{1/2}}} \\\ & \text{On rearranging it, we get: time}=n\times {{T}_{1/2}} \\\ & \Rightarrow time=5\times 164\operatorname{s} \\\ & \Rightarrow time=820s \\\ \end{aligned}$