_{35}X^{88} is an unstable isotope,which deca | vedclass

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

Question

(C) The decay process is $_{35}X^{88}$ $\xrightarrow{I} Y$ $\xrightarrow{II} _{32}Z^{84}$.Step $I$: $_{35}X^{88} \rightarrow Y + _{1}e^{0}$ (Positron

Vedclass · Accepted Answer

$Y$ and $Z$ may be isodiaphers.

Vedclass · Answer

(C) The decay process is $_{35}X^{88}$ $\xrightarrow{I} Y$ $\xrightarrow{II} _{32}Z^{84}$.Step $I$: $_{35}X^{88} \rightarrow Y + _{1}e^{0}$ (Positron emission) gives $Y = _{34}Y^{88}$.Step $II$: $_{34}Y^{88} \rightarrow _{32}Z^{84} + _{2}He^{4}$ ($\alpha$ decay).Alternatively,if $I$ is $\alpha$ decay: $_{35}X^{88} \rightarrow _{33}Y^{84} + _{2}He^{4}$,then $II$ is $\beta^-$ decay: $_{33}Y^{84} \rightarrow _{32}Z^{84} + _{-1}e^{0}$.Checking isodiaphers: Isodiaphers have the same $(N-Z)$ value.For $Y$ $(_{34}Y^{88})$: $N-Z = (88-34) - 34 = 54 - 34 = 20$.For $Z$ $(_{32}Z^{84})$: $N-Z = (84-32) - 32 = 52 - 32 = 20$.Since both have $N-Z = 20$,$Y$ and $Z$ are isodiaphers.

Similar Questions

In the following radioactive transformation given below $R$ $\xrightarrow{\alpha} X$ $\xrightarrow{\beta} Y$ $\xrightarrow{\beta} Z$; the nuclei $R$ and $Z$ are

Two nuclei which are not identical but have the same number of nucleons represent

In the treatment of cancer,which of the following isotopes is used?

Which of the following sets of nuclei represents a triad of isotones?

The triad of nuclei that is isotonic is

Vedclass Test Series

Exam Paper Generator

Online Exam Module