
A nucleus with mass number 220 initially at rest emits an \[\alpha \] particle. If the Q value of the reaction is 5.5MeV, the kinetic energy of the \[\alpha \] particle is A. 5.6MeVB. 4.4MeVC. 6.5MeVD. 5.4MeV
Answer
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Hint: First, we will calculate the mass number of the daughter nuclei that was made during the reaction. Then we will use the relationship of energy of a particle and its momentum and then apply the law of conservation of momentum to obtain the kinetic energy of the alpha particle using the Q value of the reaction.
Complete step by step solution:
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 step by step solution:
Let the nucleus with mass number be denoted by M and the daughter nuclei produced be X. Thus we have;
\[M \to X + \alpha \]
Let ${M_X}$ be the mass of daughter nuclei. The mass number of alpha particle is 4 denoted by \[{M_\alpha }\] .Thus, we have
\[{M_X} = M - {M_\alpha } \]
\[\Rightarrow {M_X} = 220 - 4 = 216 \]
Now, the Q value of the reaction is the total energy here, which is given as follows:
\[Q = {E_\alpha } + {E_X}\]
Here, \[{E_\alpha }\]is the energy of an alpha particle and \[{E_X}\] is the energy of daughter nuclei.
Now, according to the law of conservation of momentum,
\[{p_\alpha } = {p_X}\]
Here, \[{p_\alpha }\] is the momentum of the alpha particle and px is the momentum of the daughter nuclei.
Now, we know the relationship of energy and momentum as follows:
\[E = \dfrac{{{p^2}}}{{2m}}\]
Here, E is the energy of the particle while $p$ is the momentum of the particle and m is the mass of the particle.
Thus, from the law of conservation of momentum, we have
\[{{m}_{\alpha }}{{E}_{\alpha }}={{m}_{X}}{{E}_{X}} \]
\[\Rightarrow {{E}_{\alpha }}=\dfrac{{{m}_{X}}}{{{m}_{\alpha }}}Q \]
\[\therefore {{E}_{\alpha }}=\dfrac{216}{220}\times 5.5=5.4\,MeV \]
Hence, option (D) is the correct answer.
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