Chapter 29, Problem 80P

(a)

To determine

The decay scheme for ${}_{38}^{90}\text{Sr}$.

Answer to Problem 80P

The decay scheme for ${}_{38}^{90}\text{Sr}$ is ${}_{38}^{90}\text{Sr}{\to }_{39}^{90}\text{Y}{+}_{-1}^0\text{e}{+}_0^0\overline{\nu }$.

Explanation of Solution

The ${}_{38}^{90}\text{Sr}$ undergoes ${\beta }^{-}$ decay and emit an electron and changes a neutron in the nucleus to a proton. Thus, increases the charge of nucleus by $1$. The mass number of the nucleus is the same all along the process $A=90$.

From periodic table the element with atomic number $Z=39$ is calcium. And the symbol for the daughter is ${}_{39}^{90}\text{Y}$.

The ${\beta }^{-}$ decay reaction for ${}_{38}^{90}\text{Sr}$ is

    ${}_{38}^{90}\text{Sr}{\to }_{39}^{90}\text{Y}{+}_{-1}^0\text{e}{+}_0^0\overline{\nu }$

Here, $\overline{\nu }$ is an antineutrino its mass is too low, it is negligible.

Thus, the decay scheme for ${}_{38}^{90}\text{Sr}$ is ${}_{38}^{90}\text{Sr}{\to }_{39}^{90}\text{Y}{+}_{-1}^0\text{e}{+}_0^0\overline{\nu }$.

(b)

To determine

The activity of $2.0\text{ kg}$ of ${}_{38}^{90}\text{Sr}$.

Answer to Problem 80P

The activity of $2.0\text{ kg}$ of ${}_{38}^{90}\text{Sr}$ is $1.0\times {10}^{16}\text{ Bq}$.

Explanation of Solution

A sample containing ${}_{38}^{90}\text{Sr}$ with a mass of $m=2.0\text{ kg}$ and molar mass of ${}_{38}^{90}\text{Sr}$ is $M=89.9077376\text{g}/\text{mol}$ and its half-life is $T_{1/2}=28.8\text{ yr}$ . The Avogadro’s number is $N_A=6.022\times {10}^{23}\text{nuclei}/\text{mol}$.

Write the formula for half-life

    $T_{1/2}=\tau \ln 2$                                                                                            (I)

Here, $T_{1/2}$ is the half-life and $\tau$ is the time constant.

Write the formula for the number of nuclei

    $N=\frac{m{N}_A}{M}$                                                                                            (II)

Here, $N$ is the number of nuclei, $m$ is the mass, $N_A$ is the Avogadro’s number and $M$ is the molar mass.

Write the formula for the activity

    $R=\frac{N}{\tau }$                                                                                                 (III)

Here, $R$ is the activity.

Conclusion:

Substitute equation (I) and (II) in (III) to solve for $R_0$

$\begin{array}{l}R=\frac{N}{\tau }\\ R=\frac{m{N}_A\ln 2}{T_{1/2}M}\end{array}$

Substitute $2.0\text{ kg}$ for $m$, $6.022\times {10}^{23}\text{nuclei}/\text{mol}$ for $N_A$, $28.8\text{ yr}$ for $T_{1/2}$ and $89.9077376\text{g}/\text{mol}$ for $M$ in the above equation to find the value of $R$. [Note: $1\text{ yr}=3.156\times {10}^7\text{ s}$]

$\begin{array}{l}R=\frac{2.0\text{ kg}\times 6.022\times {10}^{23}\text{nuclei}/\text{mol}}{89.9077376\text{g}/\text{mol}\times 28.8\text{ yr}\times 3.156\times {10}^7\text{s}/\text{yr}}\\ R=1.0\times {10}^{16}\text{ Bq}\end{array}$

Thus, the activity of $2.0\text{ kg}$ of ${}_{38}^{90}\text{Sr}$ is $1.0\times {10}^{16}\text{ Bq}$.

(c)

To determine

The activity of ${}_{38}^{90}\text{Sr}$ after $1000\text{ yr}$.

Answer to Problem 80P

The activity of ${}_{38}^{90}\text{Sr}$ after $1000\text{ yr}$ is $3.6\times {10}^5\text{ Bq}$.

Explanation of Solution

The initial activity of ${}_{38}^{90}\text{Sr}$ is $R_0=1.0\times {10}^{16}\text{ Bq}$, then the after a period of time $t=1000\text{ yr}$ the activity $R$ is

Write the formula for the activity

    $R=R_0{e}^{-t/\tau }$                                                                                            (IV)

Here, $R$ is the activity after time $t$, $R_0$ is the activity at $t=0$ and $t$ is the time.

Conclusion:

Substitute equation (I) in (IV) to solve for $R$

$R=R_0{e}^{-t\ln 2/T_{1/2}}$

Substitute $1.0\times {10}^{16}\text{ Bq}$ for $R_0$, $28.8\text{ yr}$ for $T_{1/2}$ and $1000\text{ yr}$ for $t$ in the above equation to find the value of $R$

$\begin{array}{l}R=1.0\times {10}^{16}\text{ Bq}\times e^{-1000\text{ yr}\times \ln 2/28.8\text{ yr}}\\ R=3.6\times {10}^5\text{ Bq}\end{array}$

Thus, the activity of ${}_{38}^{90}\text{Sr}$ after $1000\text{ yr}$ is $3.6\times {10}^5\text{ Bq}$.