What is $\alpha-$,$\beta-$ and $\gamma-$decay? Write their general formulas.
(N/A) $1$. $\alpha-$decay: In this process,an unstable nucleus emits an alpha particle ($^4_2He$ nucleus). The mass number decreases by $4$ and the atomic number decreases by $2$.
General formula: $^{A}_{Z}X \rightarrow ^{A-4}_{Z-2}Y + ^{4}_{2}He + Q$
$2$. $\beta-$decay: In this process,a neutron converts into a proton (or vice versa) emitting an electron/positron and an antineutrino/neutrino. In $\beta^-$ decay,a neutron becomes a proton,emitting an electron $(e^-)$ and an antineutrino $(\bar{\nu})$.
General formula: $^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + ^{0}_{-1}e + \bar{\nu}$
$3$. $\gamma-$decay: In this process,an excited nucleus releases excess energy in the form of high-energy electromagnetic radiation (photons) without changing its mass number or atomic number.
General formula: $^{A}_{Z}X^* \rightarrow ^{A}_{Z}X + \gamma$
General formula: $^{A}_{Z}X \rightarrow ^{A-4}_{Z-2}Y + ^{4}_{2}He + Q$
$2$. $\beta-$decay: In this process,a neutron converts into a proton (or vice versa) emitting an electron/positron and an antineutrino/neutrino. In $\beta^-$ decay,a neutron becomes a proton,emitting an electron $(e^-)$ and an antineutrino $(\bar{\nu})$.
General formula: $^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + ^{0}_{-1}e + \bar{\nu}$
$3$. $\gamma-$decay: In this process,an excited nucleus releases excess energy in the form of high-energy electromagnetic radiation (photons) without changing its mass number or atomic number.
General formula: $^{A}_{Z}X^* \rightarrow ^{A}_{Z}X + \gamma$
Explore More
Similar Questions
$A$ radioactive nucleus undergoes a series of decays according to the scheme
$A \xrightarrow{\alpha} A_1 \xrightarrow{\beta} A_2 \xrightarrow{\alpha} A_3 \xrightarrow{\gamma} A_4$
If the mass number and atomic number of $A$ are $180$ and $72$,respectively,then what are these numbers for $A_4$?
$A \xrightarrow{\alpha} A_1 \xrightarrow{\beta} A_2 \xrightarrow{\alpha} A_3 \xrightarrow{\gamma} A_4$
If the mass number and atomic number of $A$ are $180$ and $72$,respectively,then what are these numbers for $A_4$?
Difficult
View Solution$_{86}A^{222} \to _{84}B^{210}$. In this reaction,how many $\alpha$ and $\beta$ particles are emitted?
Medium
View SolutionThe nucleus $_{10}^{23} Ne$ decays by $\beta^{-}$ emission. Write down the $\beta^{-}$-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
$m(_{10}^{23} Ne) = 22.994466 \; u$
$m(_{11}^{23} Na) = 22.989770 \; u$
$m(_{10}^{23} Ne) = 22.994466 \; u$
$m(_{11}^{23} Na) = 22.989770 \; u$
Medium
View Solution$A$ nucleus $_n{X^m}$ emits one $\alpha$ and one $\beta$ particle. The resulting nucleus is:
The most penetrating radiation out of the following is
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo