 Solution

ID: past-exam-of-the-mphil-in-nuclear-energy-course-of-the-university-of-cambridge/2021/paper-ne1/1/c/solution

Over

N

half-lives, the duration is

T = N t_{1/2} = N ln 2/ λ

. The time-averaged thermal power is

\overline{P} = \frac{1}{T} \int_{0}^{T} P_{0} e^{- λ t} d t = \frac{P _{0}}{λ T} (1 - e^{- λ T}) .

Since

λ T = N ln 2

and

e^{- λ T} = 2^{- N}

\overline{P} = \frac{P _{0}}{N ln 2} (1 - 2^{- N}) .

This is a time average, not the arithmetic mean of the initial and final powers. It follows directly from the assumed constant nuclear decay constant and constant recovered energy per decay.

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