A radioactive decay forms an isotope of the o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Isotopes are atoms of the same element that have the same atomic number $(Z)$ but different mass numbers $(A)$.Let the original nucleus be ${ }_{Z

Vedclass · Accepted Answer

One $\alpha$ and two $\beta$ particles

Vedclass · Answer

(C) Isotopes are atoms of the same element that have the same atomic number $(Z)$ but different mass numbers $(A)$.Let the original nucleus be ${ }_{Z}^{A} X$.After the emission of one $\alpha$ particle $({ }_{2}^{4} He)$,the atomic number decreases by $2$ and the mass number decreases by $4$:${ }_{Z}^{A} X \rightarrow { }_{Z-2}^{A-4} Y + { }_{2}^{4} He$.To return to the original atomic number $Z$,we need to increase the atomic number by $2$. This is achieved by the emission of two $\beta^-$ particles $({ }_{-1}^{0} e)$:${ }_{Z-2}^{A-4} Y + 2({ }_{-1}^{0} e) \rightarrow { }_{Z}^{A-4} Y$.Thus,the emission of one $\alpha$ particle and two $\beta$ particles results in an isotope of the original nucleus with a mass number reduced by $4$.

$A$ radioactive decay forms an isotope of the original nucleus with the emission of which of the following particles?

Similar Questions

In the disintegration series
$_{92}^{238}U \xrightarrow{\alpha} X \xrightarrow{\beta} _{Z}^{A}Y$
The values of $Z$ and $A$ respectively will be:

When an element ${}_{90}^{232}Th$ decays into ${}_{82}^{208}Pb$, the number of $\alpha$ and $\beta^{-}$ particles emitted respectively are

For the given reaction, the particle $X$ is $_6C^{11} \to _5B^{11} + \beta^+ + X$.

$A$ free neutron decays into a proton, an electron and

In a radioactive decay process, the negatively charged emitted $\beta -$ particles are

Vedclass Test Series

Exam Paper Generator

Online Exam Module