Question

The isotope ${}^{208}\mathrm{T}\mathrm{l}$ undergoes $\beta$ decay with a half-life of $3.1\mathrm{m}\mathrm{i}\mathrm{n}$.

(a) What isotope is produced by the decay?

(b) How long will it take for $99.0\mathrm{\%}$ of a sample of pure ${}^{208}\mathrm{T}\mathrm{l}$ to decay?

(c) What percentage of a sample of pure ${}^{208}\mathrm{T}\mathrm{l}$ remains un-decayed after $1.0\mathrm{h}$?

Short answer

(a) The isotope that is produced by the decay is ${}_{82}^{208}\mathrm{P}\mathrm{b}$.

(b) The time taken for$99\mathrm{\%}$of a sample of pure${}^{208}\mathrm{T}\mathrm{l}$to decay is$20.9\mathrm{m}\mathrm{i}\mathrm{n}$.

(c) The percentage of a sample of pure ${}^{208}\mathrm{T}\mathrm{l}$ that remains un-decayed after $1.0\mathrm{h}\mathrm{r}$is $1.8\cdot 10^{-4}\mathrm{\%}$.

Step-by-step solution

Concept Introduction

In nuclear physics, beta decay ($\beta -$decay) is a type of radioactive decay in which an atomic nucleus emits a beta particle (fast, energetic electron or positron) that transforms the original nuclide into an isobar of that nuclide.

The formula to calculate the ratio of the radioactive substance:

$\frac{\mathrm{N}}{{\mathrm{N}}_0}={\mathrm{e}}^{-\lambda \cdot \mathrm{t}}$

Where$\mathrm{t}$is time,$\lambda$is decay constant,${\mathrm{N}}_0$is initial concentration, and$\mathrm{N}$is the concentration at time$\mathrm{t}$.

Isotope produced by the decay

The beta decay is represented as${}_{-1}^0\mathrm{e}$.

That means the atomic number increases by$1$, and the mass number remains the same.

Now,${}_{81}^{208}\mathrm{T}\mathrm{l}$and if the atomic number increases by$1$, which is$82$, the element in the periodic table which has atomic number$82$is$\mathrm{P}\mathrm{b}$.

Therefore, the isotope is ${}_{82}^{208}\mathrm{P}\mathrm{b}$.

Decaying of the sample

From the half-life, calculate$\lambda$:

$\begin{array}{l}\lambda =\frac{\mathrm{l}\mathrm{n}(2)}{{\mathrm{t}}_{1/2}}\\ \lambda =\frac{\mathrm{l}\mathrm{n}(2)}{3.1\mathrm{m}\mathrm{i}\mathrm{n}}\\ \lambda =0.22\mathrm{m}\mathrm{i}{\mathrm{n}}^{-1}\end{array}$

The$99\mathrm{\%}$got decayed, which means the remaining amount is$1\mathrm{\%}$.

$\begin{array}{c}\frac{\mathrm{N}}{{\mathrm{N}}_0}=\frac{1}{100}\\ \frac{\mathrm{N}}{{\mathrm{N}}_0}=0.01\end{array}$

Rearranging and solving the equation:

$\begin{array}{c}\frac{\mathrm{N}}{{\mathrm{N}}_0}={\mathrm{e}}^{-\lambda \cdot \mathrm{t}}\\ 0.01={\mathrm{e}}^{-0.22\mathrm{m}\mathrm{i}{\mathrm{n}}^{-1}\cdot \mathrm{t}}\\ \mathrm{l}\mathrm{n}(0.01)=\left(-0.22\mathrm{m}\mathrm{i}{\mathrm{n}}^{-1}\cdot \mathrm{t}\right)\\ \mathrm{t}=\frac{-\mathrm{l}\mathrm{n}(0.01)}{-0.22\mathrm{m}\mathrm{i}{\mathrm{n}}^{-1}}\mathrm{t}\\ =20.9\mathrm{m}\mathrm{i}\mathrm{n}\end{array}$

Therefore, the value for time is obtained as $20.9\mathrm{m}\mathrm{i}\mathrm{n}$.

Percentage of the sample

The time given is $1\mathrm{h}\mathrm{r}=60\mathrm{m}\mathrm{i}\mathrm{n}$.

The ratio of ${\mathrm{N}}_0$ can be calculated as follows:

$\begin{array}{c}\frac{\mathrm{N}}{{\mathrm{N}}_0}={\mathrm{e}}^{-0.22\mathrm{m}\mathrm{i}{\mathrm{n}}^{-1}\cdot 60\mathrm{m}\mathrm{i}\mathrm{n}}\\ \frac{\mathrm{N}}{{\mathrm{N}}_0}=1.8\cdot 10^{-6}\cdot 100\\ \frac{\mathrm{N}}{{\mathrm{N}}_0}=1.8\cdot 10^{-4}\mathrm{\%}\end{array}$

Therefore, the value for the ratio is obtained as $1.8\cdot 10^{-4}\mathrm{\%}$.