Chapter 4, Problem 4.38P

(a)

To determine

To calculate: The activity of ${}^{90}Sr$ required to produce ${}^{50}w$ of power with $30\%$ of conversion efficiency.

Answer to Problem 4.38P

4

Activity of ${}^{90}Sr$ required is $3\times {10}^{4}Ci$ to produce ${}^{50}w$ of power in a satellite.

Explanation of Solution

Given:

Conversion efficiency= $30\%$

Power output= $50\text{W}$

Formula used: The average energy of ${}^{90}Sr$ beta particle= $\frac{1}{3}$ (maximum energy)

Required power for $30\%$ efficient energy conversion process is $50\times \frac{100}{30}=166.67watt$ .

The maximum energy of ${}^{90}Sr$ beta particle= $0.546Mev$

The average energy of ${}^{90}Sr$ beta particle= $\frac{1}{3}$ (maximum energy)

  $\begin{array}{l}=\frac{1}{3}\left(0.546\right)\hfill \\ =0.182MeV\hfill \end{array}$

As we know, ${}^{90}Sr$ is in equilibrium with 90 γ maximum energy of ${}^{90}\gamma =2.27\text{mev}$

Average energy of ${}^{90}\gamma =\frac{1}{3}\left(2.27\right)=0.76mev$

Total energy generated $=0.182+0.76=0.94mev$ per transformation.

To calculate the activity of ${}^{90}Sr$ ,

  $\begin{array}{l}=0.94mev\times 1.6\times {10}^{-13}\frac{J}{mev}\\ =1.50\times {10}^{-13}W\end{array}$

Comparing power outputs,

  $\begin{array}{l}\frac{166.67}{1.50\times {10}^{{}^{-}13}}=1.11\times {10}^{15}Bq\\ \frac{1.11\times {10}^{15}}{3.7\times {10}^{10}}=3\times {10}^{4}Ci\end{array}$

Conclusion:$3\times {10}^{4}Ci$ of activity is needed to generate a power output of $50\text{ W}$ with $30\%$ of conversion efficiency. Due to the fact that ${}^{90}Sr$ is in secular equilibrium with ${}^{90}\gamma$ Hence, we have to consider the power output of both particles that is ${}^{90}Sr$ and ${}^{90}\gamma$ .

(b)

To determine

The change in operation with use of ${}^{210}Po$ .

Answer to Problem 4.38P

It is beneficial to use ${}^{210}Po$ in the operation as only $7.2gm$ of ${}^{210}Po$ is required to produce $50w$ of power as compared to ${}^{90}Sr$ having $206.5gm$ of weight.

Hence, the weight is an important factor while designing a power source.

Explanation of Solution

Given:

T_Po=half-life of ${}^{210}Po=138$ days

T_Sr= half-life of ${}^{90}Sr=28$ years

Formula used: Average energy of ${}^{210}Po$ alpha particle = $5.4MeV$ per transformation

We know that the average energy of ${}^{210}Po$ alpha particle is $5.4MeV$ per transformation

Thus,

  $\begin{array}{l}5.4Mev\times 1.6\times {10}^{{}^{-}13}J/S\\ =8.64\times {10}^{{}^{-}13}W/Bqpower\end{array}$

But power required for $30\%$ conversion efficient process is $166.67w$

Hence,

  $\begin{array}{l}166.67/8.64\times {10}^{{}^{-}13}\\ =19.29\times {10}^{13}Bq\\ =5.2\times {10}^{3}Ci\end{array}$ of activity is required at the end of year

The activity of ${}^{210}{P}_o=\frac{A{R}_a\times T{R}_a}{A{P}_o\times T{P}_o}$

  $\begin{array}{l}=\frac{226\times 1620}{210\times (138/365)}\\ =4.61\times {10}^{3}Ci/g\end{array}$

Where, A_Ra = 266

T_Ra = 1620 yrs

A_po = Molecular weight of P_O = 210

T_Po = Half life of P_o = 138 days

Calculating activity at the start of the year according to activity at the end,

  $\begin{array}{l}{A}_0=\frac{A}{e^{-\lambda t}}\\ =\frac{5.2\times {10}^3}{e^{-(0.693/138)}\times 365}\\ =3.3\times {10}^{4}Ci\end{array}$

A = activity at the end of the year.

To calculate the weight of ²¹⁰Po compare activities of 210 Po at the start and at the end of the year.

  $\begin{array}{l}3.3\times {10}^4/4.61\times {10}^3\\ =7.2gram.\end{array}$

Now specific activity of $90Sr$ ,

  $\begin{array}{l}S.A_{S{r}_{-90}}=\frac{ARa\times TRa}{ASr\times TSr}\\ =\frac{266\times 1620}{90\times 28}\\ =145.3Ci/g\end{array}$

Where, A_Ra = 266

T_Ra = 1620 yrs

A_Sr = Molecular weight of r_S = 90

T_Sr = Half life of S_r = 25 yrs

Calculating the weight of ${}^{90}Sr$ compare activities of ${}^{90}Sr$ considering eligible decay,

  $\frac{{3.10}^4}{145.3}=206.5grams$

Conclusion:

Thus, we can say that weight required for producing $50$ watts of power over a year for ${}^{210}Po$ is $7.2$ gram and that of ${}^{90}Sr$ is $206.5$ gram. Hence, it is beneficial to use ${}^{210}Po$ as a power source of generating electrical energy.