According to this site: https://byjus.com/physics/radioactivity-gamma-decay/

Most of the time, gamma decay occurs after the radioactive nuclei have undergone an alpha or a beta decay. The alpha and beta decays leave the daughter nuclei in an excited state. From the excited state, the daughter nuclei can get back to the ground state by emitting one or more high energy gamma rays.

Call the $Q$-value the difference in the nuclear rest mass energies. From this is it true that the energy difference (assuming $Q>0$) is sometimes fully given to the kinetic energies of alpha particle, daughter nucleus, neutrino, antineutrino, but there can be situations where some part of the energy difference goes to exciting the nucleus and the rest part in kinetic energies of products?

In other words, would a more complete expression for $Q$ be $$ Q=\Delta\mathrm{KE}+\Delta H $$ where $\Delta H$ represent internal energy state of nucleus? [Considering reaction to be $\mathrm{A+B \to C^*}$ where $\mathrm C$ is in excited state]


2 Answers 2


Yes. For example, this decay scheme for cesium-137 (via, click to embiggen):

Cesium decay scheme

The beta decay has a 95% chance of putting the barium nucleus into an excited state. In this case, the primary excited state is long-lived (2.5 minutes), so you can use chemistry to separate the barium from the cesium and get a short-lived pure-gamma source. Most excited states reached by decays are short-lived, so most beta-associated gammas are effectively simultaneous with their betas.

There is also, in this system, a 5% chance that the barium nucleus is “born” in its ground state, with no associated gamma.

Note that the $Q$-value here is the ground-state-to-ground-state $Q$-value. In decays to the isomer, only about half of the $Q$-value is available for the beta decay energy.


You have understood the concept. But concept of Q value differs slightly from your approach.

If the nucleus is decaying to excited state C*, mass of C* will be slightly more than mass of C. (C* is in excited stage)

So Q value for this reaction will be less than Q value of normal reaction.

Let us say, Q value of reaction, A+B -> C is Q.

... And Q value of reaction, A+B -> C* is Q*.

Later C* decays to C by reaction C* -> C + gamma photon. Here energy of gamma photon is K.

Then Q = Q* + K

  • $\begingroup$ After a gamma decay, the nucleus will recoil (to conserve momentum and energy), so it also has some recoil energy, that you have not accounted for. $\endgroup$
    – Cross
    Commented Apr 13, 2022 at 17:28
  • $\begingroup$ @Cross : You are correct. There are few other energies I have ignored. I have assumed that A and B have negligible Kinetic energies before collision $\endgroup$ Commented Apr 13, 2022 at 22:55

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