Question
In the nuclear decay given below
$$_Z^AX \to _{Z + 1}^AY \to _{Z - 1}^{A - 4}{B^ * } \to _{Z - 1}^{A - 4}B,$$
the particles emitted in the sequence are
A.
$$\beta ,\alpha ,\gamma $$
B.
$$\gamma ,\beta ,\alpha $$
C.
$$\beta ,\gamma ,\alpha $$
D.
$$\alpha ,\beta ,\gamma $$
Answer :
$$\beta ,\alpha ,\gamma $$
Solution :
Alpha particles are positively charged particles with charge $$+2e$$ and mass $$4\,m.$$ Emission of an $$\alpha $$-particle reduces the mass of the radionuclide by 4 and its atomic number by 2. $$\beta $$-particles are negatively charged particles with rest mass as well as charge same as that of electrons. $$\gamma $$-particles carry no charge and mass.
Radioactive transition will be as follows
$$\eqalign{
& _Z^AX \to _{Z + 1}^AY + \beta _{ - 1}^0 \cr
& _{Z + 1}^AY \to _{Z - 1}^{A - 4}\beta + \alpha _2^4 \cr
& _{Z + 1}^{A - 4}\beta \to _{Z - 1}^{A - 4}\beta + \gamma _0^0 \cr} $$