Chapter 24, Problem 107A

Interpretation Introduction

Interpretation:

Amount of Flourine-18 left from $\text{15}\text{g}$ after $\text{8}\text{hr}$ is to be estimated.

Concept introduction:

In a nuclear reaction, unstable nuclide disintegrates into stable nuclide with emission of various radiations spontaneously. It is called radioactivity of nucleus. Emission of rays like alpha, beta and gamma rays occur in the process which changes the identity of nucleus.

During a nuclear change, an element undergoes change in its atomic mass or atomic number or both by emission of charged particles such as beta particle, alpha particle. Beta particle is one that has unit negative charge and alpha particle is one that has 2 unit positive charges and is equivalent to helium atom ${}_2^4\text{He}$. Emission of proton, beta particle, alpha particle causes a change in atomic number of element.

Answer to Problem 107A

The remaining amount of radioactive Flourine-18 is $0.72403\text{g}$.

Explanation of Solution

Given information:

Elapsed time is $\text{8}\text{hr}$, mass of Flourine-18 sample is $\text{15}\text{g}$ and half life is $\text{1}\text{.82951}\text{hr}$.

The formula to calculate the number of half lives ( $n$ ) is as follows:

$n=\frac{t}{T}$

Where,

$n$ is number of half lives.

$t$ is the elapsed time.

$T$ is the half life of radioactive nucleus.

Substitute $\text{168}\text{hr}$ for $t$, $\text{65}\text{.9239}\text{hr}$ for $T$ in above equation.

$\begin{array}{c}n=\frac{t}{T}\\ =\frac{\text{8}\text{hr}}{\text{1}\text{.82951}\text{hr}}\\ =4.37275\end{array}$

Number of half lives equal to $4.37275$.

The formula to calculate the remaining amount of radioactive element is as follows:

$\text{N}={\text{N}}_{\text{0}}{\left(\frac{1}{2}\right)}^n$

Where,

$n$ is number of half lives.

${\text{N}}_{\text{0}}$ is the initial amount of radioactive element.

$\text{N}$ is the remaining amount of radioactive element.

Substitute $4.37275$ for $n$, $\text{15}\text{g}$ for ${\text{N}}_{\text{0}}$ in above equation.

$\begin{array}{c}\text{N}={\text{N}}_{\text{0}}{\left(\frac{1}{2}\right)}^n\\ =\left(15\text{g}\right){\left(\frac{\text{1}}{\text{2}}\right)}^{4.37275}\\ =0.72403\text{g}\end{array}$

The remaining amount of radioactive Flourine-18 is $0.72403\text{g}$.

Conclusion

The remaining amount of radioactive Flourine-18 is $0.72403\text{g}$.