Various ways of generating electricity from small radioactive power sources have been long known.

The Voyager space probes launched in the 1970s use thermoelectric generation from plutonium sources to generate ~100W of power to keep the probes running when they go very far from the sun. Some small devices have used other sources and other conversion mechanisms to prove power for very low power applications like pacemakers where long lasting power that doesn't need to be replaced is an advantage.

But recent claims suggest a breakthrough where devices powered by carbon-14 could replace the batteries in mobile phones (Nano diamond Batteries web page highlights stories claiming their produces could "revolutionise mobile phone batteries").

These claims seem like uncritical hype but should be easy to check.

Ignoring engineering issues (like conversion efficiency) how does the energy density of a power source based on C-14 compare to a mobile phone battery?

(I'm guessing a simple calculation based on the radioactive properties of C-14 could give a decent answer and that any battery experts can tell us what a Li-ion battery achieves so we can compare them. Also, nuclear power sources are not "batteries" but they are compact power sources so please don't be distracted by the terminology).

  • 3
    $\begingroup$ The isotopic power of C-14 is 0.001305 W/g. That should be the theoretical upper limit even for an ideal system. $\endgroup$
    – Loong
    Commented Feb 18, 2023 at 19:27
  • 1
    $\begingroup$ Some energy would be carried away by electron antineutrinos. For given thermal energy, the efficiency of energy conversion by Seeback effect thermocouples in space probes (usually with $\ce{^{238}Pu}$ with $\pu{0.57 W/g}$) is reportedly like 10%. There are prototypes using the Stirling engine with up to 40% efficiency, but with doubts about the long term reliability due moving parts. $\endgroup$
    – Poutnik
    Commented Feb 18, 2023 at 19:40
  • 1
    $\begingroup$ @Poutnik neutrinos are excluded from the value I gave above; so it's the energy available to the decay heat problem. $\endgroup$
    – Loong
    Commented Feb 18, 2023 at 19:51
  • 2
    $\begingroup$ @matt_black When calculating myself: The half life of C-14 is 5700 a, which gives you a decay constant of 3.853E-12/s or 2.321E+12 Bq/mol or 1.657E+11 Bq/g. The maximum beta energy is 0.15648 MeV, but the mean beta energy is only 0.04945 MeV = 7.923E-15 J. So the isotopic power is 1.657E+11 Bq/g * 7.923E-15 J = 1.313E-03 W/g. The value given by my old Nuclides 2000 is 1.3050E-03 W/g. $\endgroup$
    – Loong
    Commented Feb 18, 2023 at 20:54
  • 1
    $\begingroup$ @Poutnik I'm happy to be corrected on the appropriate terminology to use for the comparison with conventional batteries. I suspect the relevant one is gravimetric power density in this case. Correct me if I'm wrong. $\endgroup$
    – matt_black
    Commented Feb 18, 2023 at 23:08

1 Answer 1


The maximum theoretical power density of isotopically pure $\ce{^14C}$ can be estimated from the nuclear decay data of $\ce{^14C}$ as follows.

The half-life of $\ce{^14C}$ is $T_{1/2}=5.70\cdot 10^3\ \mathrm a$ 1. Thus, the corresponding decay constant is $$\tau=\frac{\ln2}{T_{1/2}}=3.9\cdot 10^{-12}\ \mathrm s^{-1}$$ and the molar activity is $$A_\mathrm M=\tau N_\mathrm A=2.3\cdot 10^{12}\ \mathrm{Bq\ mol^{-1}}$$ Since the molar mass of $\ce{^14C}$ is $M=14.0\ \mathrm{g\ mol^{-1}}$, the specific activity of $\ce{^14C}$ is $$a=\frac{A_\mathrm M}M=1.7\cdot 10^{11}\ \mathrm{Bq\ g^{-1}}$$

The energy of the beta decay of $\ce{^14C}$ is $E=156.48\ \mathrm{keV}$ 1. However, the beta spectrum is continuous because the total energy of the decay process is divided between the electron, the antineutrino, and the recoiling nuclide. The mean beta energy only amounts to $E_\mathrm\beta=49.45\ \mathrm{keV}$ 1.

C-14 beta spectrum

Since $$E_\mathrm\beta=49.45\ \mathrm{keV}=7.923\cdot 10^{-15}\ \mathrm J$$ the power density (so-called isotopic power) of $\ce{^14C}$ is $$e_\mathrm\beta=aE_\mathrm\beta=1.3\cdot 10^{-3}\ \mathrm{W\ g^{-1}}$$

How does this compare to a typical mobile phone battery?

An iPhone 14 pro has a Li-ion battery with capacity of about 3,200 mAh. At a voltage of 3.6 V, this gives a capacity of about 11.5 Wh.

It weighs just over 200g. While the power consumption varies a lot depending on what the phone is doing these have been estimated to peak at over 4W for some other smartphones. We can estimate the average power consumption by noting that a typical iPhone needs to be recharged every one or two days. 11.5 Wh over a 48h time period suggests an average power consumption of about 240mW.

To provide the average power (ignoring conversion efficiency and any other engineering issues) this would require about 185g of C-14. But to provide the peak power (say 4W) would require more than 3kg of C-14 which is 15 times heavier than the total weight of an existing iPhone 14 pro.

This would be truly a revolution in mobile phone batteries, but not the good sort.


1 ICRP Publication 107. Nuclear Decay Data for Dosimetric Calculations. Ann. ICRP 2008, 38 (3).

  • $\begingroup$ Overall efficiency after accounting for the isotope separation is pretty lousy… $\endgroup$
    – Jon Custer
    Commented Feb 19, 2023 at 13:55
  • 2
    $\begingroup$ By way of comparison: 100 g C-14 is about 1.7E+13 Bq which approximately corresponds to the annual C-14 production of a CANDU reactor. The global atmospheric inventory of C-14 from natural sources is about 1.4E+15 Bq or 8.4 kg. $\endgroup$
    – Loong
    Commented Feb 19, 2023 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.