Question
In one $$\alpha $$ and 2 $$\beta $$-emissions
A.
mass number reduces by 2
B.
mass number reduces by 6
C.
atomic number reduces by 2
D.
atomic number remains unchanged
Answer :
atomic number remains unchanged
Solution :
The $$\alpha $$-particle can be represented as $$_2H{e^4}$$ and $$\beta $$-particle as $$_{ - 1}{\beta ^0}.$$ So, after emission of one $$\alpha $$-particle the mass number of resultant nucleus decreases by 4 unit and atomic number by 2 unit. Similarly, after emission of one $$\beta $$-particle the atomic number increases by 1 unit keeping its mass number same. So, according to reaction (assuming $$_Z{X^A}$$ the initial nucleus)
$$\eqalign{
& _Z{X^A}{ \to _{Z - 2}}{Y^{A - 4}}{ + _2}H{e^4}\,\,\left( {\alpha - {\text{particle}}} \right) \cr
& {\text{and}}\,{\,_{Z - 2}}{Y^{A - 4}}{ \to _Z}{X^{A - 4}} + 2\left( {_{ - 1}{\beta ^0}} \right)\,\,\left( {2\beta - {\text{particle}}} \right) \cr} $$
So, by one $$\alpha $$ and two $$\beta $$-emissions the atomic number remains unchanged. i.e. formation of isotopes takes place.