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# In an $\alpha$ decay, the kinetic energy of $\alpha$ particles is $48MeV$ and $Q$ value of the reaction is $50MeV$. The mass number of the mother nucleus is $X$. Find the value of $X/25$. (Assume that daughter nucleus is in the ground state)A) $2$ B) $4$ C) $6$ D) $8$

Last updated date: 13th Sep 2024
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Hint: Before proceeding with our solution, let us discuss the various terms we will use. Alpha decay is a radioactive decay in which a nucleus emits an alpha particle (which is nothing but an ionised helium nucleus) and thereby transforms into a different atomic nucleus, with a mass number that is reduced by four and an atomic number that is reduced by two. The Q value for a reaction is the amount of energy absorbed or released during the nuclear reaction and can be determined from the masses of reactants and products. Q values affect reaction rates.

Formula Used:
${{K}_{\alpha }}=\dfrac{{{m}_{d}}}{{{m}_{\alpha }}+{{m}_{d}}}Q$

Complete step by step solution:
As we know that alpha decay causes changes in the atomic number and mass number of the mother nucleus, mathematically shown as

${}_{x}{{A}^{Y}}\to {}_{X-2}{{B}^{Y-4}}+{}_{2}H{{e}^{4}}$ where $A$ is the mother nucleus, $B$ is the daughter nucleus, $He$ is the alpha particle and the subscripts denote the atomic numbers of the respective nuclei and the superscripts denote the mass number of the respective nuclei.

We know the relation between the kinetic energy of the alpha particle and the total energy of the reaction is given as

${{K}_{\alpha }}=\dfrac{{{m}_{d}}}{{{m}_{\alpha }}+{{m}_{d}}}Q$ where ${{K}_{\alpha }}$ is the kinetic energy of the alpha particle, $Q$ is the total energy change of the reaction, ${{m}_{d}}$ is the mass of the daughter nucleus and ${{m}_{\alpha }}$ is the mass of the alpha particle.

We have been given that kinetic energy of the alpha particle $({{K}_{\alpha }})=48MeV$

Total energy change in the reaction $(Q)=50MeV$
From the reaction of the alpha decay of a mother nucleus given above, we can say that
Mass of the daughter nucleus $({{m}_{d}})=X-4$
We know that mass of an alpha particle $({{m}_{\alpha }})=4$
Substituting the values given to us, we get
\begin{align} & 48=\dfrac{X-4}{X-4+4}\times 50 \\ & \Rightarrow \dfrac{X-4}{X}=\dfrac{48}{50} \\ \end{align}
Cross multiplying, we get
\begin{align} & 50X-200=48X \\ & \Rightarrow 2X=200 \\ & \Rightarrow X=100 \\ \end{align}
But we have to find the value of $X/25$ which will be $100/25=4$

Hence option (B) is the correct answer.

Note: We were not provided with the mass of the daughter nucleus; we had to find it using the mass of the mother nucleus and the mass of an alpha particle. Also, keep in mind that we were asked the value of the mass of the mother nucleus divided by four and not the mass itself. Students almost always choose the wrong option in such cases, even after getting the right answer.