
In an $\alpha $-decay the kinetic energy of $\alpha $ particle is 48 MeV and the Q-value of the reaction is 50MeV. The mass number of the mother nucleus is: (Assume that daughter nucleus is in ground state)
A. 9
B. 100
C.104
D. None of these
Answer
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Hint: In this question we have to apply the concept of $\alpha $ decay. We have to use the mass of the products forming and the mass of the nucleus before the reaction. We have to use the formula of the Q-value of reaction and with that we can find the value of the mass number of the mother nucleus.
Complete step by step answer:We know that in an $\alpha $ decay reaction the mass number decreases by value of 4 and the atomic number decreases with the value of 2. To represent it in an equation it is given by:
\[{}_Y^XB \to {}_{Y - 2}^{X - 4}C + \alpha + Q\]
Where, B and C are the mother and daughter nucleus respectively, and we have considered the mass number of the mother nucleus as X.
Here we can see the change in the atomic number and mass number of the mother and daughter nucleus.
Now if we consider the formula of Q-value of the reaction, it is given by:
$Q = {E_\alpha }\left( {\dfrac{{{M_\alpha } + {M_C}}}{{{M_C}}}} \right)$
Now it is given to use that $Q = 50MeV\,{\rm{and }}{E_\alpha } = 48MeV$,
We will put these values and the mass of the alpha particle and the Q-Value in the formula to find out the Mass of the mother nucleus.
Putting all the values we get:
$\begin{array}{l}
Q = {E_\alpha }\left( {\dfrac{{4 + X - 4}}{{X - 4}}} \right)\\
\Rightarrow Q = {E_\alpha }\left( {\dfrac{X}{{X - 4}}} \right)
\end{array}$
We have put the values of the mass number of the alpha particle and the daughter nucleus.
Going further:
$\begin{array}{l}
Q = {E_\alpha }\left( {\dfrac{X}{{X - 4}}} \right)\\
\Rightarrow 50 = 48\left( {\dfrac{X}{{X - 4}}} \right)\\
\Rightarrow 50X - 200 = 48X\\
\Rightarrow 2X = 200\\
\Rightarrow X = 100
\end{array}$
Now that we have calculated X, which we considered the mass of the mother nucleus in the beginning we know that the correct answer is 100 and the correct option is (B).
Note:In such types of problems the student’s understanding of the radioactive decay reactions is very important. Also, the relations should be well remembered by a student, such as of Q-value. While calculating the units should be in the same unit systems.
Complete step by step answer:We know that in an $\alpha $ decay reaction the mass number decreases by value of 4 and the atomic number decreases with the value of 2. To represent it in an equation it is given by:
\[{}_Y^XB \to {}_{Y - 2}^{X - 4}C + \alpha + Q\]
Where, B and C are the mother and daughter nucleus respectively, and we have considered the mass number of the mother nucleus as X.
Here we can see the change in the atomic number and mass number of the mother and daughter nucleus.
Now if we consider the formula of Q-value of the reaction, it is given by:
$Q = {E_\alpha }\left( {\dfrac{{{M_\alpha } + {M_C}}}{{{M_C}}}} \right)$
Now it is given to use that $Q = 50MeV\,{\rm{and }}{E_\alpha } = 48MeV$,
We will put these values and the mass of the alpha particle and the Q-Value in the formula to find out the Mass of the mother nucleus.
Putting all the values we get:
$\begin{array}{l}
Q = {E_\alpha }\left( {\dfrac{{4 + X - 4}}{{X - 4}}} \right)\\
\Rightarrow Q = {E_\alpha }\left( {\dfrac{X}{{X - 4}}} \right)
\end{array}$
We have put the values of the mass number of the alpha particle and the daughter nucleus.
Going further:
$\begin{array}{l}
Q = {E_\alpha }\left( {\dfrac{X}{{X - 4}}} \right)\\
\Rightarrow 50 = 48\left( {\dfrac{X}{{X - 4}}} \right)\\
\Rightarrow 50X - 200 = 48X\\
\Rightarrow 2X = 200\\
\Rightarrow X = 100
\end{array}$
Now that we have calculated X, which we considered the mass of the mother nucleus in the beginning we know that the correct answer is 100 and the correct option is (B).
Note:In such types of problems the student’s understanding of the radioactive decay reactions is very important. Also, the relations should be well remembered by a student, such as of Q-value. While calculating the units should be in the same unit systems.
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