The number of alpha- and beta- particles emit | vedclass

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

Question

(A) The radioactive decay process is given by: $_{90}E^{232} \rightarrow _{86}G^{220} + x(\alpha) + y(\beta)$.Number of $\alpha$ particles $(x)$ is ca

Vedclass · Accepted Answer

$3$ $\alpha$ and $2$ $\beta$

Vedclass · Answer

(A) The radioactive decay process is given by: $_{90}E^{232} \rightarrow _{86}G^{220} + x(\alpha) + y(\beta)$.Number of $\alpha$ particles $(x)$ is calculated by the change in mass number: $x = \frac{232 - 220}{4} = \frac{12}{4} = 3$.Number of $\beta$ particles $(y)$ is calculated using the change in atomic number: $90 = 86 + 2x - y$,which gives $y = 86 + 2(3) - 90 = 86 + 6 - 90 = 2$.Therefore,$3$ $\alpha$ particles and $2$ $\beta$ particles are emitted.

The number of $\alpha-$ and $\beta-$ particles emitted when a radioactive element $_{90}E^{232}$ changes into $_{86}G^{220}$ will be

Similar Questions

In the following nuclear reaction,identify $x$: $_{29}Cu^{64} \to _{28}Ni^{64} + x$

Consider the radioactive disintegration ${}_{82}A^{210}$ $\rightarrow B$ $\rightarrow C$ $\rightarrow {}_{82}D^{206}$. The sequence of emission can be

$^{27}_{13}A$ is a stable element. $^{29}_{13}A$ is expected to disintegrate by:

What happens when an $\alpha$-particle is emitted?

Who developed the study of radioactivity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module