# Is radioactive decay independent of the environment?

## Main Question or Discussion Point

Is the rate of radioactive decay fixed or does the environment have any impact eg would the rate of decay be the same in a low or very high gravitational field (in both cases measured from the viewpoint of the radioactive material)?

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vanhees71
Gold Member
2019 Award
In some special cases you can have huge effects. One example is bound $\beta$ decay, where the electron decays into a bound atomic state rather than a free (scattering) state. Of course it can do so only if the corresponding atomic state is not occupied by an electron since electrons are fermions. A famous example is Rh 187 which as a neutral atom decays with a half-life of about $42 \cdot 10^9 \, \text{y}$, making it apparently to a perfect clock to measure the age of astrophysical objects by measuring the abundance ratio of Rh and Os (it's decay product). However, when the Rh is ionized, then bound-state $\beta$ decay can take place. Then the half-life becomes of the order of $10 \; \text{y}$, i.e., a 9 orders of magnitude smaller value. In fact the half-life of ionized Rh 187 has been measured at GSI in Darmstadt (Germany) to be $33 \;\text{y}$. For a very nice review by Fritz Bosch, one of the scientists involved in these measurements, see

http://www.euroschoolonexoticbeams.be/site/files/nlp/LNP651_contrib5.pdf

DrChinese
Gold Member
Is the rate of radioactive decay fixed or does the environment have any impact eg would the rate of decay be the same in a low or very high gravitational field (in both cases measured from the viewpoint of the radioactive material)?
Gravity does affect the radioactive 1/2 life in the sense that time appears to "slow down" when enough mass is present.

Staff Emeritus
2019 Award
In some special cases you can have huge effects.
For example, some nuclei decay by electron capture. If you are looking at fully stripped nuclei (no electrons), these decays don't happen.

vanhees71
Yep, that's a crossing-symmetric reaction to my example of bound $\beta$ decay :-)).