Discussion
\(_{1}H^{3}\) nuclei undergoes beta decay at a rate of 1.27 × 10\(^{17}\) particles/hour and produce 21 calories of heat per hour in a medium. The average energy of the beta particle emitted is :
Answer: D
The average beta energy is about 4.3 KeV.
Convert the heat to joules and divide by the number of decays:
\[ \begin{aligned} \bar{E} &= \frac{21 \times 4.186\ \text{J}}{1.27 \times 10^{17}} = 6.92 \times 10^{-16}\ \text{J} \\ &= \frac{6.92 \times 10^{-16}}{1.6 \times 10^{-19}}\ \text{eV} \approx 4.3\ \text{keV} \end{aligned} \]
As an average it lies well below the 18.6 keV end point of tritium decay, as it must, since the beta spectrum is continuous. The larger options would need more heat or fewer decays.
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