A nucelus with atomic number `Z` and neutron number `N` undergoes two decay processes. The result is a nucleus with atomic number `Z-3` and neutron `N-1`. Which decay processes took place?

Options

A. Two `beta^(bar)` decays
B. Two `beta^(+)` decays
C. An `alpha` decay and a `beta^(bar)` decays
D. `An `alpha` decay and a `beta^(+)` decays

Answer

d

Explanation

Two protons and two neutrons are lost in an `alpha`-decays, so `Z` and `N` each decreases by `2`. A `beta^(bar)` decay changes a proton to a neutron, so `Z` decreases by `1` and `N` increases by `1`. The net result is `Z` decreases by `3` and `N` decreases by `1`.
`._(Z)^(A)Xoverset(alpha-decay)(rarr)._(Z-2)^(A-4)Yoverset(beta-decay)(rarr)._(Z-3)^(A-4)Z`
Initially, number of neutrons `N_i=(A-Z)`
Now, number of neutrons `N_(f)=A-4-Z+3=N_(i)-1`.