The nucleus 2310Ne decays by β emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m (2310Ne) = 22.994466 u


m (2311Ne) = 22.089770 u.


For β- decay,the number of proton increases by 1.The reaction is given as:


Here the electron masses gets cancelled as Na has one more electron than Ne hence,


Q = (22.994466-22.089770)u × c2


= 0.00469 × 931.5 MeV


= 4.37 MeV


The maximum k.E of the emitted electrons are comparable to the Q value; Hence the maximum K.E of the electrons emitted will be = 4.37 MeV


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