Answer
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Hint :Use the relation between the $ Q $ - value, kinetic energy of the alpha particle and the mass number $ A $ of the mother nucleus. The $ Q $ - value relation of alpha decay given by, $ K{E_\alpha } = \left( {\dfrac{{A - 4}}{A}} \right)Q $ where, $ K{E_\alpha } $ is the kinetic energy of the $ \alpha $ - particle, $ A $ is the mass number of the mother nucleus.
Complete Step By Step Answer:
We know that the equation of decay is given by, $ {}_Z^AM \to {}_{Z - 2}^{A - 4}D + {}_2^4He $
is the $ M $ mother nucleus and is the $ D $ daughter nucleus at ground state. $ {}_2^4He $ is an alpha particle.
We know, that the $ Q $ - value of an $ \alpha $ -decay is given by, $ Q = \dfrac{A}{{A - 4}}K{E_\alpha } $ where, $ K{E_\alpha } $ is the kinetic energy of the $ \alpha $ - particle, $ A $ is the mass number of the mother nucleus. $ Q $ - value is the energy released in the disintegration process.
On changing the sides of $ K{E_\alpha } $ and $ Q $ we get, $ K{E_\alpha } = \left( {\dfrac{{A - 4}}{A}} \right)Q $
We have given here that the $ Q $ - value of the reaction is $ 50MeV $ . So, $ Q = 50MeV $ . The kinetic energy of the $ \alpha $ -particle is $ 48MeV $ or $ K{E_\alpha } = 48MeV $ .
Hence putting the values we get,
$ \Rightarrow 48 = \left( {\dfrac{{A - 4}}{A}} \right)50 $
On simplifying we get,
$ 48A = (A - 4)50 $
Or, $ 50A - 48A = 200 $
Or, $ 2A = 200 $
Hence, the value of $ A $ mass number of the mother nucleus is, $ A = 100 $
Hence, Option ( B) is correct.
Note :
Since, the daughter nucleus is in ground states then it cannot decay further, So, we have calculated for decay of one $ \alpha $ particle.
The general formula for $ Q $ - value of alpha particle disintegration is given by, $ Q = \left( {\dfrac{{{M_d} + {M_\alpha }}}{{{M_d}}}} \right)KE $
Where, $ {M_d} $ is the mass of the daughter nucleus, $ {M_\alpha } $ is the mass of the alpha particle.
Hence, the general relation is in terms of their masses but it is noticed that the ratio of their masses is the same as the ratio of their mass numbers. So, we use the formula containing the mass numbers.
Complete Step By Step Answer:
We know that the equation of decay is given by, $ {}_Z^AM \to {}_{Z - 2}^{A - 4}D + {}_2^4He $
is the $ M $ mother nucleus and is the $ D $ daughter nucleus at ground state. $ {}_2^4He $ is an alpha particle.
We know, that the $ Q $ - value of an $ \alpha $ -decay is given by, $ Q = \dfrac{A}{{A - 4}}K{E_\alpha } $ where, $ K{E_\alpha } $ is the kinetic energy of the $ \alpha $ - particle, $ A $ is the mass number of the mother nucleus. $ Q $ - value is the energy released in the disintegration process.
On changing the sides of $ K{E_\alpha } $ and $ Q $ we get, $ K{E_\alpha } = \left( {\dfrac{{A - 4}}{A}} \right)Q $
We have given here that the $ Q $ - value of the reaction is $ 50MeV $ . So, $ Q = 50MeV $ . The kinetic energy of the $ \alpha $ -particle is $ 48MeV $ or $ K{E_\alpha } = 48MeV $ .
Hence putting the values we get,
$ \Rightarrow 48 = \left( {\dfrac{{A - 4}}{A}} \right)50 $
On simplifying we get,
$ 48A = (A - 4)50 $
Or, $ 50A - 48A = 200 $
Or, $ 2A = 200 $
Hence, the value of $ A $ mass number of the mother nucleus is, $ A = 100 $
Hence, Option ( B) is correct.
Note :
Since, the daughter nucleus is in ground states then it cannot decay further, So, we have calculated for decay of one $ \alpha $ particle.
The general formula for $ Q $ - value of alpha particle disintegration is given by, $ Q = \left( {\dfrac{{{M_d} + {M_\alpha }}}{{{M_d}}}} \right)KE $
Where, $ {M_d} $ is the mass of the daughter nucleus, $ {M_\alpha } $ is the mass of the alpha particle.
Hence, the general relation is in terms of their masses but it is noticed that the ratio of their masses is the same as the ratio of their mass numbers. So, we use the formula containing the mass numbers.
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