Chapter 19, Problem 19.96QP

(a)

Interpretation Introduction

Interpretation:

Equation for the given decay process has to be written.

Concept Introduction:

The total number of protons and neutrons in the product and in reactant side must be equal.

The total number of nuclear charges in the product and reactant side must be equal.

${}_Z^{A}X$, where X is the element, A is the mass number and Z is the atomic number ${}_2^{4}He$ has mass number 4 and atomic number 2.

(b)

Interpretation Introduction

Interpretation:

The power output at zero time and at time 10year when initially 1.0mg of $\text{Pu-238}$ is used has to be determined.

(c)

Interpretation Introduction

When $\text{Pu-238}$ isotope decays by emitting an alpha particle. The balanced equation is

${}_{94}^{238}Pu\to {}_{92}^{234}U+{}_2^{4}He$

At time $=0$, the number of $\text{Pu-238}$ can be calculated as,

$\left(1.0\times {10}^{-3}g\right)\frac{1mol}{238g}\times \frac{6.023\times {10}^{23}atoms}{1mol}=2.53\times {10}^{18}atoms$

$NumberPu-238atomspresentatt=0,=2.53\times {10}^{18}atoms$

Use,

$\lambda =\frac{0.693}{t_{1/2}}$

$\begin{array}{l}\lambda =\frac{0.69}{86yr}=8.058\times {10}^{-3}y{r}^{-1}\\ \lambda =8.058\times {10}^{-3}y{r}^{-1}\times \frac{1yr}{365d}\times \frac{1d}{24h}\times \frac{1h}{3600s}=2.55\times {10}^{-10}{s}^{-1}\\ \\ Use,\\ rate=\lambda N_0\\ rate=2.55\times {10}^{-10}{s}^{-1}\times 2.53\times {10}^{18}atoms=6.45\times {10}^{8}decay/s\end{array}$