If a nucleus P converts into a nucleus Q by t | vedclass

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

Question

(A) Let the nucleus $P$ be represented as $_{Z}^{A}P$, where $A$ is the mass number and $Z$ is the atomic number.An alpha particle is represented as $

Vedclass · Accepted Answer

Isotopes

Vedclass · Answer

(A) Let the nucleus $P$ be represented as $_{Z}^{A}P$, where $A$ is the mass number and $Z$ is the atomic number.An alpha particle is represented as $_{2}^{4}\alpha$ and a $\beta^{-}$ particle (electron) is represented as $_{-1}^{0}e$.The decay process is: $_{Z}^{A}P \rightarrow _{Z'}^{A'}Q + 1(_{2}^{4}\alpha) + 2(_{-1}^{0}e)$.Conservation of mass number: $A = A' + 4 + 2(0) \Rightarrow A' = A - 4$.Conservation of atomic number: $Z = Z' + 2 + 2(-1) \Rightarrow Z = Z' + 2 - 2 \Rightarrow Z = Z'$.Since the atomic number $Z$ remains the same $(Z = Z')$, the nuclei $P$ and $Q$ have the same number of protons.Nuclei with the same atomic number but different mass numbers are called isotopes.

If a nucleus $P$ converts into a nucleus $Q$ by the decay of one alpha particle and two $\beta^{-}$ particles, then the nuclei $P$ and $Q$ are

Similar Questions

$\beta$-decay means emission of electron from

$A$ nucleus of an element ${}_{84}X^{202}$ emits an $\alpha$-particle first,a $\beta$-particle next,and then a gamma photon. The final nucleus formed has an atomic number:

In the uranium series,the decay constant of the parent nuclide is $\lambda$. What is the decay constant of the stable end product of this series?

$37$ Rutherford equals

$A$ radioactive nucleus with $Z$ protons and $N$ neutrons emits an $\alpha$-particle,$2\beta^-$-particles,and $2$ gamma rays. The number of protons and neutrons in the nucleus left after the decay,respectively,are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module