
When a Radioactive decay can form an isotope of the original nucleus what are the particles emitted?
A.One $\alpha $ and four $\beta $
B. One $\alpha $ and two $\beta $
C. One $\alpha $ and one $\beta $
D. Four $\alpha $ and one $\beta $
Answer
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Hint: Radioactive decay is a process in which an unstable nucleus loses energy by radiation. An alpha particle is a helium nucleus. When an alpha particle is emitted the atomic number of the parent element is reduced by 2 and the mass number is reduced by 4. When a beta particle is emitted the atomic number increases by one and the mass number remains unchanged Isotopes are atoms of same element having same mass number but different atomic number
Complete step by step answer:
Radioactive decay is a process in which an unstable nucleus loses energy by radiation. It can be of three types: alpha decay beta decay or gamma decay.
An alpha particle is a helium nucleus. When an alpha particle is emitted the atomic number of the parent element is reduced by 2 and the mass number is reduced by 4. When a beta particle is emitted the atomic number increases by one and the mass number remains unchanged
Suppose the parent nuclei is represented as $X_Z^A$, Where $A$ is the mass number and $Z$is the atomic number
Now let us suppose an alpha particle is emitted from this nucleus. Then the daughter nuclei will be given by
$X_Z^A\xrightarrow{\alpha }Y_{Z - 2}^{A - 4}$
Isotopes are atoms of the same element having the same atomic number but different mass number.
From the above equation we can see that the atomic number is $2$ less than the atomic number of the parent nucleus. We know that in beta emission the atomic number is increased by one thus if two beta particles are emitted then we can get daughter nuclei with the same atomic number $Z$ . Thus ,
$Y_{Z - 2}^{A - 4}\xrightarrow{{2\beta }}Z_Z^{A - 4}$
Now we have the isotope of the original nucleus. Hence one $\alpha $ and two $\beta $ particles should be emitted in this case. So the answer is option B
Note:
There are two types of beta decay. ${\beta ^{^ - }}$decay and ${\beta ^{^ + }}$ decay. In ${\beta ^{^ - }}$ decay neutron in the parent nucleus is converted into proton, electron and electron antineutrino. Here electrons are emitted and the atomic number of the parent nucleus is increased by $1$. Whereas in ${\beta ^{^ + }}$ decay a proton in the parent nuclei is converted into positron, neutron and electron neutrino. Here positron is emitted and the atomic number of the parent nuclei is decreased by $1$. In this question we considered the ${\beta ^{^ - }}$ decay where the atomic number of the parent nucleus increases by $1$.
Complete step by step answer:
Radioactive decay is a process in which an unstable nucleus loses energy by radiation. It can be of three types: alpha decay beta decay or gamma decay.
An alpha particle is a helium nucleus. When an alpha particle is emitted the atomic number of the parent element is reduced by 2 and the mass number is reduced by 4. When a beta particle is emitted the atomic number increases by one and the mass number remains unchanged
Suppose the parent nuclei is represented as $X_Z^A$, Where $A$ is the mass number and $Z$is the atomic number
Now let us suppose an alpha particle is emitted from this nucleus. Then the daughter nuclei will be given by
$X_Z^A\xrightarrow{\alpha }Y_{Z - 2}^{A - 4}$
Isotopes are atoms of the same element having the same atomic number but different mass number.
From the above equation we can see that the atomic number is $2$ less than the atomic number of the parent nucleus. We know that in beta emission the atomic number is increased by one thus if two beta particles are emitted then we can get daughter nuclei with the same atomic number $Z$ . Thus ,
$Y_{Z - 2}^{A - 4}\xrightarrow{{2\beta }}Z_Z^{A - 4}$
Now we have the isotope of the original nucleus. Hence one $\alpha $ and two $\beta $ particles should be emitted in this case. So the answer is option B
Note:
There are two types of beta decay. ${\beta ^{^ - }}$decay and ${\beta ^{^ + }}$ decay. In ${\beta ^{^ - }}$ decay neutron in the parent nucleus is converted into proton, electron and electron antineutrino. Here electrons are emitted and the atomic number of the parent nucleus is increased by $1$. Whereas in ${\beta ^{^ + }}$ decay a proton in the parent nuclei is converted into positron, neutron and electron neutrino. Here positron is emitted and the atomic number of the parent nuclei is decreased by $1$. In this question we considered the ${\beta ^{^ - }}$ decay where the atomic number of the parent nucleus increases by $1$.
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