Question

An element $E$ loses one $\alpha$ and two $\beta$-particles in three successive stages. The resulting element will be:

• A. an isobar of $E$
• B. an isotone of $E$
• C. an isotope of $E$
• D. $E$ itself

Answer 1

Answer: (C), (D)

The correct options are
C $E$ itself
D an isotope of $E$

Let the mass number of $E$ is $A$ and atomic number $=Z$

since $\alpha {=}_2^{4}He,\beta {=}_{-1}^0\beta$

one alpha and two beta decays of E are represented as:

${}_Z^{A}E\stackrel{{}_2^{4}He}{⟶}{}_{Z-2}^{A-4}X\stackrel{{}_{-1}^0\beta }{⟶}{}_{Z-1}^{A-4}Y\stackrel{{}_{-1}^0\beta }{⟶}{}_Z^{A-4}Z$

Thus the resulting element $Z$ has same atomic number as $E(=Z)$ but different mass number, therefore the decay gives an isotope of $E$ or same element $E$.