Cobalt-60 atomic battery

Problem

A sample of cobalt-60 ($T_{1/2} = 5{,}27\ \text{years}$) has a mass of $0{,}50\ \text{g}$. The $\beta^-$ decay releases on average $0{,}31\ \text{MeV}$ of kinetic energy per beta particle, plus two gamma photons of $1{,}33\ \text{MeV}$ and $1{,}17\ \text{MeV}$. Calculate: (a) the initial number of atoms; (b) the initial activity; (c) the initial thermal power, knowing that the total energy released per decay is $E_{\text{tot}} \approx 2{,}81\ \text{MeV}$.

Solution

(a) Initial number of atoms. The molar mass of cobalt-60 is $\approx 60\ \text{g/mol}$. With Avogadro’s number $N_A = 6{,}022\cdot 10^{23}$:

$N = \frac{m}{M}\,N_A = \frac{0{,}50}{60}\cdot 6{,}022\cdot 10^{23} \approx \ev{5{,}02\cdot 10^{21}\ \text{atoms}}$

(b) Initial activity. The decay constant is $\lambda = \ln 2 / T_{1/2}$, with $T_{1/2}$ in seconds ($1\ \text{year} \approx 3{,}156\cdot 10^7\ \text{s}$):

$\lambda = \frac{0{,}693}{5{,}27\cdot 3{,}156\cdot 10^{7}} = \frac{0{,}693}{1{,}663\cdot 10^{8}} \approx 4{,}17\cdot 10^{-9}\ \text{s}^{-1}$

The initial activity is $A = \lambda N$:

$A = 4{,}17\cdot 10^{-9}\cdot 5{,}02\cdot 10^{21} \approx \ev{2{,}09\cdot 10^{13}\ \text{Bq} \approx 565\ \text{Ci}}$

(having used $1\ \text{Ci} = 3{,}7\cdot 10^{10}\ \text{Bq}$).

(c) Initial thermal power. Each decay releases $E_{\text{tot}} \approx 2{,}81\ \text{MeV} = 2{,}81\cdot 1{,}6\cdot 10^{-13}\ \text{J}$. The power is the activity times the energy per decay:

$P = A\cdot E_{\text{tot}} = 2{,}09\cdot 10^{13}\cdot 2{,}81\cdot 1{,}6\cdot 10^{-13} \approx \ev{9{,}4\ \text{W}}$

This is the order of magnitude of the thermal power exploited in radioisotope power sources.

Collegamenti

Argomenti: Fisica nucleare Concetti: Decadimento radioattivo · Equivalenza massa-energia