Question

A radioactive nucleus ${}_Z{X}^A$ emits $3\alpha$-particles and $5\beta$-particles. The ratio of number of neutron, protons in the product nucleus will be :-

• A. $\frac{A-Z-12}{Z-6}$
• B. $\frac{A-Z}{Z-1}$
• C. $\frac{A-Z-11}{Z-1}$
• D. $\frac{A-Z-12}{Z-1}$

Answer 1

The correct option is C $\frac{A-Z-11}{Z-1}$

The new mass number after emission of $3$ alpha particles will be $A-3\times 4=A-12$

the new atomic number after emission of $3$ alpha particles will be $Z-3\times 2=Z-6$

and after emission of $5$ beta particles, this will become $Z-6+5\times 1=Z-1$

so the number of $neutrons$ in the new nucleus will be $A-12-(Z-1)=A-Z-11$

and the number of $protons$ in the new nucleus will be $(Z-1)$

the required ration will be $\frac{n}{p}=\frac{A-Z-11}{Z-1}$