Question

Polonium ${(}_{84}{\text{Po}}^{210})$ emits ${}_2{\alpha }^4$ particles and is converted into lead ${(}_{82}{\text{Pb}}^{206})$. This reaction is used for producing electric power in a space mission. ${\text{Po}}^{210}$ has a half-life of $138.6\text{days}$. Assuming an efficiency of $10\%$ of the thermoelectric machine, how much ${\text{Po}}^{210}$ is required to produce $1.2\times {10}^7\text{J}$ of electric energy per day at the end of $693\text{days}$?

$\text{Given masses of the nuclei}{\text{Po}}^{210}=209.98264\text{amu},{\text{Pb}}^{206}=205.97440\text{amu},{\alpha }^4=4.00260\text{amu},1\text{amu}\times c^2=931\text{MeV}\text{Avogadro number=}6\times {10}^{23}.$

• A. $10\text{g}$
• B. $320\text{g}$
• C. $640\text{g}$
• D. $100\text{g}$

Answer 1

The correct option is B $320\text{g}$

The given reaction is represented as,

${}_{84}{\text{Po}}^{210}\to {}_{82}{\text{Pb}}^{206}+{}_2{\text{He}}^4$

$\text{Mass defect :}$

$\Delta m=209.98264\text{amu}-(205.97440\text{amu}+4.00260\text{amu})$

$\Rightarrow \Delta m=0.00564\text{amu}$

$\text{Equivalent energy}$

$E=0.000564\times 931\text{MeV}$

$\Rightarrow E=5.25\text{MeV}=5.25\times 1.6\times {10}^{-13}\text{J}$

$\Rightarrow E=8.4\times {10}^{-13}\text{J}$

$\text{As efficiency is 10\%, equivalent energy}$

$E=0.1\times 8.4\times {10}^{-13}=0.84\times {10}^{-13}\text{J}$

Given that, the total amount of energy required,

$E^{\prime }=1.2\times {10}^7\text{J}$

Therefore, Number of reactions required per day

$N=\frac{E^{\prime }}{E}=\frac{1.2\times {10}^7}{0.84\times {10}^{-13}}=\frac{1}{7}\times {10}^{21}$

This is the activity of the sample.

$\text{Activity}A=\lambda N_{atom}$

$A=\frac{\ln 2}{t_{1/2}}{N}_{atom}$

$\Rightarrow \frac{1}{7}\times {10}^{21}=\frac{0.693}{138.6}{N}_{atom}$

$\Rightarrow N_{atom}=\frac{200}{7}\times {10}^{21}$

We know that,

$N_{atom}=\frac{m}{At.wt.}{N}_A$

$\Rightarrow \frac{200}{7}\times {10}^{21}=\frac{m}{210}\times 6\times {10}^{23}$

$\Rightarrow m=10\text{g}$

Given that,

$t=693\text{days},t_{1/2}=138.6\text{days}$

$\text{Number of half-lives,}n=\frac{t}{t_{1/2}}$

$\Rightarrow n=\frac{693}{138.6}=5$

$\text{Initial mass of}{m}_0\text{required},$

$\frac{m}{m_0}=\frac{N_{atom}}{N_0}={\left(\frac{1}{2}\right)}^n$

$\Rightarrow \frac{m}{m_0}={\left(\frac{1}{2}\right)}^5=\frac{1}{32}$

$\Rightarrow m_0=32m=320\text{g}$

Hence, option $\left(B\right)$ is correct.