Is decay heat proportional to half life?

I know different isotopes decay at different rates. The rate for a given isotope is given by its half life. I know heat is produced by decay events.
I know it is bad form to put 2 questions in one box but I am going to do it because they are small ones.

Question 1: Is decay heat produced in a given time per mole of isotope proportional to half life? One would think more rapid decay would produce more heat but I cannot find that written down.

Question 2: Is it possible for heat generated per unit time to increase? I could imagine that if an isotope decayed to a series of isotopes with shorter half-lives (which produced more heat?) that heat produced for a given quantity of mass might increase before it decreased.

The rate of decay (activity) $\dfrac{dN}{dt}$ is proportional to the number of undecayed nuclei $N$.

This is usually written as

$\dfrac{dN}{dt} \propto N \Rightarrow \dfrac {dN}{dt} = - \lambda N$

where $\lambda$ is the decay constant which is related to the half life $\tau$ as $\lambda\, \tau = \ln 2$

So $\dfrac{dN}{dt} = - \dfrac{\ln 2}{\tau} N$

The energy released from a nuclear decay is called the Q-value and so the rate of heat generation is

$\dfrac{dN}{dt} Q = - \dfrac{\ln 2}{\tau} N Q$

So for a given number of undecayed nuclei $N$ ($\propto$ number of moles) the rate of production of heat is inversely proportional to the half life.

You could imagine a very long lived parent nuclei with a Q-value for decay of $Q_{\rm p}$ which decay into much shorter lived daughter nuclei with Q-value for decay $Q_{\rm d}$ with the decay of a daughter nucleus releasing more energy than that of a parent nucleus $Q_{\rm d} >Q_{\rm p}$.

Suppose initially there are no daughter nuclei but after a time an equilibrium situation is set up so that the rates of decay of the parent nuclei and the daughter nuclei are the same.

So $\dfrac{dN_{\rm p}}{dt} = \dfrac{dN_{\rm d}}{dt}$.

This will mean that the rate of heat production from the decay of the daughter nuclei $\left ( \dfrac{dN_{\rm d}}{dt}\, Q_{\rm d} \right )$ will be greater than the rate of heat production from the decay of the parent nuclei $\left ( \dfrac{dN_{\rm p}}{dt}\, Q_{\rm p} \right )$ which would represent an increase in the rate of heat production.

Update in response to a comment.

If you have a parent to daughter decay, starting with no daughter nuclei present, then the graphs of activity will be of the form depending on the relative half lives of the parent $\tau_{\rm p}$ and daughter $\tau_{\rm d}$.

The example I used above was for secular equilibrium where $\tau_{\rm p} \gg \tau_{\rm d}$

To get the rate of energy release one would need to multiply the activity by the appropriate Q-value for the decay.
So it is that product which will determine which decay mode is releasing energy at the greatest rate.

The mathematics relating to radioactive decay chains was first published by Harry Bateman in 1910 but you find that subsequent derivations and explanations easier to comprehend?
Googling Bateman equation will provide you with many links.

• Thanks for your answer. If the daughter nuclei are shorter lived how would the rate of decay for them ever be the same as the rate for the parent? I thought decay rate was an intrinsic property. – Willk Aug 11 '17 at 1:20
• @Will, Decay per mole of isotope is intrinsic. But there are not equal amounts of P and D. Initially D=0. Only as much D will accumulate until the rates (over the entire amount) are equal. – BowlOfRed Aug 11 '17 at 4:09

When you use the word "proportional" you are implying a comparison of some kind (though admittedly one or both sides of the comparison may be hypothetical), and we need to ask 'Just what is being compared?' before we can answer your question.

The problem is complicated because in general two sources with different halflives don't have the same energy release per decay, but I'm going to assume we can deal with that. I'm also going to use lifetimes rather than halflives because I think more clearly in base $e$.

Heat generated now by same-population samples of different life times

Assume we have two samples each containing $N$ atoms of two isotopes with the same average energy release $q$ per decay but differing lifetimes ($\tau_l> \tau_s$ where 'l' is for long and 's' is for short).

In a time $\delta t$ that is short compared to either life time, the heat generated is proportional to the current rate of decays which is inversely related to the life time. \begin{align} Q_\text{now} &= N q \exp \left( \frac{\delta t}{\tau} \right) \\ \\ &\approx N q \left(1 + \frac{\delta t}{\tau} \right) \end{align} so the ratio is \begin{align} R_\text{now} &= \frac{ Q_{\text{now},l} }{ Q_{\text{now},s} } \\ \\ &= \frac{1 + \frac{\delta t}{\tau_l}}{1 + \frac{\delta t}{\tau_s}} \\ \\ &\approx 1 - \frac{\delta t}{\tau_l} \frac{\tau_s}{\delta t} \\ &= 1 - \frac{\tau_s}{\tau_l} \;. \end{align}

In short the heat evolved by the longer lived sample is less than that evolved by the short-lived sample. If the ratio of halflives is large enough this can approach being inversely proportional to life time.

Heat generated much later by (currently) same-population samples of different life times

In this case we wait $\Delta T$ considerably larger than either lifetime before sampling. In this case the surviving population $$N_\text{later} = N \exp\left( -\frac{\Delta T}{\tau} \right) \;,$$ and the heat evolved at that later time is \begin{align} Q_\text{now} &\approx N \exp\left( -\frac{\Delta T}{\tau} \right) q \left(1 + \frac{\delta t}{\tau} \right) \;, \end{align} which makes the ratio \begin{align} R_\text{later} &\approx \exp \left[ -\Delta T \left( \frac{1}{\tau_l} - \frac{1}{\tau_s} \right) \right] \left( 1 - \frac{\tau_s}{\tau_l} \right) \;, \end{align} of the two surviving term the exponential dominates, so changing the order to remove the minus sign we get $$R \propto \exp \left[ \Delta T \left( \frac{1}{\tau_s} - \frac{1}{\tau_l} \right) \right] \;,$$ and it is clear that heat evolved by the longer lived sample exceeds that of the shorter lived sample.

This shouldn't come as a surprise because we effectively assumed that the two samples had the same total energy available and the short-lived one developed more at the start, so the clearly the long-lived one has more available later on.