I recently been studying different radioactive elements. The most difficult part is find the output of these elements in the form of kWh.

Is there a database or chart of these elements' output?

And most importantly does anyone know the kWh of $1~ kg$ of $\ce{^99Tc}$?

  • $\begingroup$ Welcome to chemistry.SE! If you had any questions about the policies of our community, you can ‎visit the help center or take a ‎‎tour of the website‎. $\endgroup$
    – M.A.R.
    Commented Jun 11, 2015 at 17:39
  • $\begingroup$ I don't quite follow the question. kWh is a unit of energy. Are you asking what the total energy released if 1 kg of technetium-99m transitions to technetium-99 emitting mostly gamma rays? (I assume you mean Tc-99m because Tc-99 has an extremely long half-life, but correct me if I'm wrong) $\endgroup$ Commented Jun 11, 2015 at 18:12
  • $\begingroup$ According to Nuclides 2000, the isotopic power of Tc-99 is only 8.6E−06 W/g; the value for Tc-99m is about 4.4E+03 W/g. $\endgroup$
    – user7951
    Commented Jun 11, 2015 at 18:15
  • $\begingroup$ I'm trying to calculate the energy out put of TC 99 like 1 kg uranium 235 puts out 50,000 kWh $\endgroup$
    – user46173
    Commented Jun 11, 2015 at 18:16
  • $\begingroup$ Can you clarify if you are talking about Tc-99 or Tc-99m. $\endgroup$
    – bon
    Commented Jun 11, 2015 at 18:41

1 Answer 1


$\ce{^{99m}Tc}$ is a nuclear isomer of $\ce{^{99}Tc}$, which is excited by 142.6 keV. Therefore, when one atom decays to $\ce{^{99}Tc}$, it releases 142.6 keV, mostly as gamma radiation, but also some internal conversion occurs. Therefore, for 1 kg of $\ce{^{99m}Tc}$ (10.1 mol):

$$ (142.6\ \mathrm{keV})(10.1\ \mathrm{mol})N_A = 8.673\times 10^{29}\ \mathrm{eV} = 139\ \mathrm{GJ} = 38.6\ \mathrm{MWh} $$

In general, the total energy released by a nuclear reaction in a given mass is simply the decay energy (commonly known as the Q value) multiplied by the number of atoms. The Q value is simply the difference in mass between the starting material and final product converted to energy:

$$Q = (m_\mathrm{i}-m_\mathrm{f})c^2$$

Q values are easy to just look up, though. (Wolfram|Alpha is a good place)

Of course, depending on the application, all this energy is not accessible or useful. If one were trying to make an RTG, an enormous amount of shielding would be needed to turn the gamma rays into heat, since they penetrate matter very well, where things like Pu-238 work better because they release mostly alpha particles, which are easily absorbed, delivering their kinetic energy as heat.


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.