$8_4 mathrm{Po}^{210}$ decays with $alpha$-particles to ${ }_{82} mathrm{~Pb}^{206}$ with a half-life of 138.4 days. If 1 g of ${ }_{84} mathrm{Po}^{210}$ is placed in a sealed tube, the amount of helium(in $mathrm{cm}^3$ ) that will accumulate in 69.2 days will be:

It is proposed to use <img alt="_{84} \mathrm{PO}^{210} \rightarrow\:\: _{82}\mathrm{Pb}^{210}+\, _{2}\mathrm{He}^4" src="/latex-image/?_%7B84%7D%20%5Cmathrm%7BPO%7D%5E%7B210%7D%20%5Crightarr

It is proposed to use $_{84} \mathrm{PO}^{210} \rightarrow\:\: _{82}\mathrm{Pb}^{210}+\, _{2}\mathrm{He}^4$ to produced $2 \mathrm{~KW}$ of electric power in a generator. The half life of polonium $\left(_{84} \mathrm{Pd}^{210}\right)$ is $138.6$ days. Assuming efficiency of the generator to be $10 \%$ how many gm of polonium are required per day:

$\text { Masses of nuclei } \mathrm{PO}^{210}=209.98264\, \mathrm{amu}$
$\mathrm{Pb}^{206}=205.97440\, \mathrm{amu}, \mathrm{He}^4=4.00260 \, \mathrm{amu} \text {. }$
and $\mathrm{1 \mathrm{amu}=931 \: \mathrm{Mev}}$ .
Option: 1
$12.0 \, \mathrm{gm}$

Option: 2
$14.40 \, \mathrm{gm}$

Option: 3
$12.20 \, \mathrm{gm}$

Option: 4
$144\, \mathrm{gm}$

Answers (1)

$\mathrm{\Delta m=m\left ( _{84}po^{210} \right )-\left [ m\left ( _{82}Pb^{210} \right )+m\left (_{2}He^{4} \right )\right ]}$
$= 0.00564 \, \mathrm{amu}$
$\mathrm{ \Delta E=\Delta m c^2=0.00564 \times 931=5.25\, \mathrm{MeV} }$ .

$\mathrm{decay \: constant\, \lambda= \frac{0.693}{t_{1 / 2}}=\frac{0.693}{138.6}=0.005 /day}$ .
If $\mathrm{M\, \mathrm{gm}}$ of $\mathrm{po}$ required per day, the number of nuclei in $\mathrm{M\, gm}$ is $\mathrm{\frac{6 \times 10^{23}}{210} \times \mathrm{M}=\mathrm{N}}$

$\Rightarrow\mathrm{}$ Rate of decay $\mathrm{= \lambda N}$
$\mathrm{ \text { Energy produced / day } =\lambda N \times 5.25 \times 1.6 \times 10^{-19} \times 10^{6}J}$
$\mathrm{ =2 \times 10^3 \mathrm{W} \times 24 \times 60 \times 60}$

$\mathrm{ \Rightarrow M=14.4 \mathrm{gm}}$ .

Since the efficiency is $\mathrm{ 10 \% \Rightarrow}$ The amount required is $\mathrm{ 144\, \mathrm{gm}}$ .

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