Answer:
The answer to the type of radioactive decay that isotope with a high value of n/z will undergo is beta decay.
Step-by-step explanation:
An atom comprises three subatomic particles; The electrons, the protons, and the neutron. The n/z is the ratio of the number of the neutron to the number of protons in the same atom. Atoms of the first twenty elements have an average of the n/z ratio of 1, apart from the hydrogen, which is nil, Helium, and Be. The effect of n/z to the stability of the atom is very pronounced and the atom may become very unstable when it is very high. That means that the number of neutrons in that atom is higher than the amount of the proton. Thus, there tends to be the conversion of the excess neutrons to protons until there is an equilibrium. The transformation of neutrons to the positive intranuclear subatomic proton, therefore, requires Beta decay. This is so because the beta radiation that will be emitted are in form of electrons, which will render the atom more positively charged, and this will nullify the effect of the neutrons depending on the amount of electrons that are released.