Question

Strontium-90 is one of the most hazardous products of atomic weapons testing because of its long half-life(${\mathbf{t}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{28}\mathbf{.}\mathbf{1} \mathbf{years}$) and its tendency to accumulate in the bone.

1. Write nuclear equations for the decay of ${}^{\mathbf{90}}\mathbf{Sr}$via the successive emission of two electrons (beta particles).
2. The atomic mass of ${}^{\mathbf{90}}\mathbf{Sr}$is 89.9073 u and that of role="math" localid="1663338677065" ${}^{\mathbf{90}}\mathit{Z}\mathbf{r}$is 89.9043 u. Calculate the energy released per ${}^{\mathbf{90}}\mathbf{Sr}$ atom, in MeV, in decaying to ${}^{\mathbf{90}}\mathit{Z}\mathbf{r}$.
3. What will be the initial activity of 1.00 g of ${}^{\mathbf{90}}\mathbf{Sr}$released into the environment, in disintegrations per second?
4. What activity will material from part (c) show after 100 years.

Short answer

1. ${}_{38}{}^{90}Sr\to {}_{40}{}^{90}Zr+2{}_{+1}{}^0{e}^{-1}+2\overline{\nu }$
2. The energy released is $2.79 MeV$
3. The initial activity is $4.68\times {10}^{12}{s}^{-1}$
4. The activity after 100 years is$4.13\times {10}^{11}{s}^{-1}$

Step-by-step solution

Nuclear equation representing beta emission by  Sr90

In a nucleus when the number of neutrons is excess than the number of protons, $\beta$decay takes place. When $\beta$decay takes place, a neutron is converted to a proton and a electron.

The high-energy electron emission is known as the $\beta$particle. The mass number of the daughter nucleus remains same, while the atomic number of the daughter nucleus increases by 1. An antineutrino is emitted along with the $\beta$particle.

The balanced equation that represents electron emission by ${}^{\text{90}}\text{Sr}$is:

${}_{38}{}^{90}Sr\to {}_{40}{}^{90}Zr+2{}_{+1}{}^0{e}^{-1}+2\overline{\nu }$

Energy released due to electron emission

The mass difference caused due to emission of two particles is calculated as follows:

$\begin{array}{rcl}\Delta m& =& m\left({}_{40}{}^{90}Zr\right)-m\left({}_{38}{}^{90}Sr\right)\\ & =& 89.9043u-89.9073u\\ & =& 0.003 u\end{array}$

The energy equivalent to a mass difference of $=-0.003u\times 931.5 MeV=-2.79 MeV$

The energy released = $2.79 MeV$

Initial activity

$\begin{array}{rcl}k& =& \frac{0.693}{t_{1/2}}\\ & =& \frac{0.693}{28.1 years }\\ & =& \frac{0.693}{28.1\times 365\times 24\times 3600s}\\ & =& 7.8\times {10}^{-10} s^{-1}\end{array}$

Number of atoms in 1.00 g$=\frac{1.00 g}{90 g/mol}\times \frac{6.023\times {10}^{23}atoms}{1mol}=0.06\times {10}^{23}atoms$

$\begin{array}{rcl}A& =& kN\\ & =& 7.8\times {10}^{-10}\times 0.06\times {10}^{23}{s}^{-1}\\ & =& 4.68\times {10}^{12}{s}^{-1}\end{array}$

Activity after 100 years

$\begin{array}{rcl}{A}_t& =& A_0{e}^{-kt}\\ & =& 4.68\times {10}^{12}\times e^{-7.8\times {10}^{-10}y{r}^{-1}\times 100yr}\\ & =& 4.13\times 10{}^{11} atoms\end{array}$