
If the speed of \[\alpha \] particle is v, find recoil speed of daughter nuclei.
Answer
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Hint: When the mass number of an atom is very high, it has more number of neutrons and hence the atom becomes radioactive. We will apply the law of conservation of linear momentum to find the recoil speed of the daughter nuclei. Here, the parent nuclei’s velocity is zero as the nucleus is at rest.
Complete answer:
Here, when the mass number of an atom is very high, it means that its atomic number is also very high. Now, because of this, there will be more neutrons and hence the atom becomes unstable. Thus, the atom becomes radioactive and starts decaying and the atom breaks into a daughter nuclei and alpha particle. In our case, the parent nuclei have a mass number of A and it is at rest, hence its velocity is zero. The mass number of an alpha particle is 4.
Thus when the parent nucleus emits an α particle its mass number decreases by 4. The mass of the daughter nucleus is (A−4). Let the velocity of the alpha particle be $u$ and the velocity of daughter nuclei of $v$. Now, when the reaction processes, the total momentum of the reaction is conserved. Thus, the conservation of momentum is as follows:
\[
A(0)+4u=(A-4)v \\
\therefore v=\dfrac{4u}{A-4}
\]
Hence, the recoil speed of daughter nuclei is $\dfrac{4u}{A-4}$.
Note: The mass number of the daughter nuclei is always less than that of the parent nuclei. Also, if the summation of the mass number of daughter nuclei and the alpha particle is less than that of the parent nuclei, then it means that some of the remaining mass is converted into heat energy which is the basis of a nuclear bomb.
Complete answer:
Here, when the mass number of an atom is very high, it means that its atomic number is also very high. Now, because of this, there will be more neutrons and hence the atom becomes unstable. Thus, the atom becomes radioactive and starts decaying and the atom breaks into a daughter nuclei and alpha particle. In our case, the parent nuclei have a mass number of A and it is at rest, hence its velocity is zero. The mass number of an alpha particle is 4.
Thus when the parent nucleus emits an α particle its mass number decreases by 4. The mass of the daughter nucleus is (A−4). Let the velocity of the alpha particle be $u$ and the velocity of daughter nuclei of $v$. Now, when the reaction processes, the total momentum of the reaction is conserved. Thus, the conservation of momentum is as follows:
\[
A(0)+4u=(A-4)v \\
\therefore v=\dfrac{4u}{A-4}
\]
Hence, the recoil speed of daughter nuclei is $\dfrac{4u}{A-4}$.
Note: The mass number of the daughter nuclei is always less than that of the parent nuclei. Also, if the summation of the mass number of daughter nuclei and the alpha particle is less than that of the parent nuclei, then it means that some of the remaining mass is converted into heat energy which is the basis of a nuclear bomb.
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