98/43 Tc undergoes gamma decay. How would you determine the resulting nucleus?

1 Answer
Jul 3, 2016

Interesting question since #"_43^98 Tc# is a #beta^-#-emitter.


Technetium (Tc) is a nuclide that only has unstable isotopes. Each isotope decays in its own way. You can use a table of nuclides to find the mode of decay of a certain isotope.

#"_43^98 Tc# decays to #""_44^98 Ru# (Ruthenium) by emitting #beta^-#particles. Following this #beta# decay it does emit gamma rays (745 and 652 keV).

So the statement that #"_43^98 Tc# undergoes gamma decay is actually not true unless they mean that the isotope is in a short lived excited state. When you add energy to a nucleus it can get into an excited state. The nucleus will emit the excess energy as gamma rays. This doesn't change the isotope itself.

Pure gamma decay is only found in metastable isotopes such as #"^(99m)Tc# which is a longer lived excited state. With a half life of 6 hours it decays to #"^99Tc# by emitting gamma rays. In general gamma decay does not change the nucleus, it only changes the energy level of the nucleus.