In the nuclear reaction _{85}X^{297} to Y + 4 | vedclass

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

Question

(C) The given nuclear reaction is $_{85}X^{297} 	o Y + 4\alpha$.An alpha particle is represented as $_2He^4$.Therefore,the reaction can be written as

Vedclass · Accepted Answer

$_{77}Y^{281}$

Vedclass · Answer

(C) The given nuclear reaction is $_{85}X^{297} 	o Y + 4\alpha$.An alpha particle is represented as $_2He^4$.Therefore,the reaction can be written as $_{85}X^{297} 	o Y + 4(_2He^4)$.To find the atomic number $(Z)$ and mass number $(A)$ of $Y$,we apply the conservation of charge and mass number.For the mass number $(A)$: $297 = A + 4 	imes 4 \implies 297 = A + 16 \implies A = 297 - 16 = 281$.For the atomic number $(Z)$: $85 = Z + 4 	imes 2 \implies 85 = Z + 8 \implies Z = 85 - 8 = 77$.Thus,the resulting nucleus $Y$ is $_{77}Y^{281}$.

In the nuclear reaction $_{85}X^{297} \to Y + 4\alpha$,$Y$ is:

Similar Questions

$A$ sample of a radioactive nucleus $A$ disintegrates to another radioactive nucleus $B$,which in turn disintegrates to some other stable nucleus $C$. Plot a graph showing the variation of the number of atoms of nucleus $B$ versus time:
(Assume that at $t=0$,there are no $B$ atoms in the sample)

Which of the following can penetrate through an aluminum sheet of $20 \, cm$ thickness?

$A$ radioactive element $_{90}X^{238}$ decays into $_{83}Y^{222}$. The number of $\beta$-particles emitted is:

In the uranium radioactive series,the initial nucleus is $_{92}U^{238}$ and the final nucleus is $_{82}Pb^{206}$. When the uranium nucleus decays to lead,the number of $\alpha$-particles emitted will be

Who discovered radioactivity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module