Chapter 21, Problem 21.114QA

Interpretation Introduction

To find:

The element that will be present in the greatest amount after $1 year.$

Answer to Problem 21.114QA

Solution:

$Pb$ will be present in the greatest amount after $1 year$.

Explanation of Solution

1) Concept:

We are using the first order radioactive decay equation to find the ratio $N_t/N_0$ for each process given. Larger the ratio, larger will be the value of $N_t$, which means it will be present in the greatest amount after time $t$.

2) Formula:

i) $\ln \frac{Nt}{N_0}=-0.693\frac{t}{t_{\frac{1}{2}}}$      

where, $N_t$ is the value of substance remained after time $t$, $N_0$ is the initial amount of substance, $t$ is time of the process, and $t_{\frac{1}{2}}$ is half-life of the substance.

3) Given:

i) Reaction series and its respective half-life is given below:

$Bi \stackrel{{\rm A}}{\to } Ti \stackrel{{\rm B}}{\to } Pb \stackrel{{\rm B}}{\to } Bi\to$

$t_{\frac{1}{2}}=20 min 1.3 min 20 yr 5d$

ii) $t=1 year$

4) Calculations:

i) Ratio $\frac{N_t}{N_0}$  for $Bi \stackrel{{\rm A}}{\to } Ti, t_{\frac{1}{2}}=20 min$:

As half-life is given in min, we need to convert $t=1 year$ to min.

$t=1 year \times \frac{365 days}{1 year}\times \frac{24 hour}{1 day}\times \frac{60 min}{1 hour}=525600 min$

$\ln \frac{N_t}{N_0}=-0.693\frac{t}{t_{\frac{1}{2}}}=-0.693\times \frac{525600 min}{20 min}=-18212.04$

$\frac{N_t}{N_0}=e^{\left(-18212.04\right)}=0$

ii) Ratio $\frac{N_t}{N_0}$ for $Ti \stackrel{{\rm B}}{\to } Pb, t_{\frac{1}{2}}=1.3 min$:

$\ln \frac{N_t}{N_0}=-0.693\frac{t}{t_{\frac{1}{2}}}=-0.693\times \frac{525600 min}{1.3 min}=-280185.2$

$\frac{N_t}{N_0}=e^{(-280185.2)}=0$

iii) Ratio $\frac{N_t}{N_0}$ for $Pb \stackrel{{\rm B}}{\to } Bi, t_{\frac{1}{2}}=20 years$:

$\ln \frac{N_t}{N_0}=-0.693\frac{t}{t_{\frac{1}{2}}}=-0.693\times \frac{1 year}{20 years}=-0.03465$

$\frac{N_t}{N_0}=e^{(-0.03465)}=0.97$

iv) Ratio $\frac{N_t}{N_0}$ for $Bi\to , t_{\frac{1}{2}}=5 days$:

As half-life is given in days, we need to convert $t=1 year$ to days.

$t=1 year \times \frac{365 days}{1 year}=365 days$

$\ln \frac{N_t}{N_0}=-0.693\frac{t}{t_{\frac{1}{2}}}=-0.693\times \frac{365 days}{5 days}=-50.589$

$\frac{N_t}{N_0}=e^{\left(-50.589\right)}=1.07{\times }^{-22}$

The ratio $\frac{N_t}{N_0}$ for all four nuclei given in series is calculated. For $Pb$, we got the largest value of this ratio. Larger the ratio $\frac{N_t}{N_0}$, larger will be the value of $N_t$. Therefore, $Pb$ will be present in the greatest amount after $1 year.$

Conclusion:

Larger the ratio $\frac{N_t}{N_0}$, larger will be the value of $N_t$ after time $t$. So, ratio $\frac{N_t}{N_0}$ of all nuclei is calculated and compared.