
If \[M{\text{ }}\left( {A,{\text{ }}Z} \right)\], \[{M_p}\], and \[{M_n}\] denote the masses of the nucleus ${}_Z^AX$ proton and neutron respectively in units of u \[\left( {1u{\text{ }} = {\text{ }}931.5MeV/{C^2}} \right)\] and BE represents its bonding energy is MeV, then-
(A) $M(A,Z) = Z{M_p} + (A - Z){M_n} - BE$
(B) $M(A,Z) = Z{M_p} + (A - Z){M_n} + \frac{{BE}}{{{c^2}}}$
(C) $M(A,Z) = Z{M_p} + (A - Z){M_n} - \frac{{BE}}{{{c^2}}}$
(D) $M(A,Z) = Z{M_p} + (A - Z){M_n} + BE$
Answer
218.1k+ views
Hint Einstein’s mass energy relation is \[E = m{c^2}\], here, E is the binding energy, m is the mass defect and c is the speed of light. Mass defect is the mass anomaly in calculating mass in a nucleus. This mass defect times \[{c^2}\] is equivalent to Binding Energy (BE).
Complete step-by-step solution
Binding energy: The neutrons and protons in a stable nucleus are held together by nuclear forces and energy is needed to pull them infinitely apart, this energy is called binding energy.
$BE = \Delta m \times {c^2}$
Mass defect: It is found that the mass of the nucleus is always less than the sum of the masses of its constituent nucleons in free state, this difference in mass is called mass defect. It is given by,
$\Delta m = \{ Z{M_p} + (A - Z){M_n}\} - M(A,Z)$
Where,
\[M{\text{ }}\left( {A,{\text{ }}Z} \right)\], \[{M_p}\], and \[{M_n}\] are mass of nucleus, proton and neutron respectively. and Z is mass and atomic number.
Thus, by substituting the value of ∆m in BE, we get
$\begin{gathered}
BE = \{ Z{M_p} + (A - Z){M_n}\} - M(A,Z) \times {c^2} \\
\frac{{BE}}{{{c^2}}} = \{ Z{M_p} + (A - Z){M_n}\} - M(A,Z) \\
M(A,Z) = \{ Z{M_p} + (A - Z){M_n}\} - \frac{{BE}}{{{c^2}}} \\
\end{gathered} $
Hence, the correct option is C
Note The binding energy is expressed in J when mass defect is in kg. And for converting from mass to energy, one needs to multiply \[{c^2}\] to it. To convert from Energy to mass, it has to be divided by \[{c^2}\]. If mass defect is in amu then binding energy is 931 MeV times \[\Delta m\].
In subjects like general relativity and quantum mechanics, the factor \[{c^2}\] that differentiates between mass and energy is ignored; the two quantities are considered equivalent. Just like how the number 1 is ignored after being multiplied to some arbitrary quantity.
Complete step-by-step solution
Binding energy: The neutrons and protons in a stable nucleus are held together by nuclear forces and energy is needed to pull them infinitely apart, this energy is called binding energy.
$BE = \Delta m \times {c^2}$
Mass defect: It is found that the mass of the nucleus is always less than the sum of the masses of its constituent nucleons in free state, this difference in mass is called mass defect. It is given by,
$\Delta m = \{ Z{M_p} + (A - Z){M_n}\} - M(A,Z)$
Where,
\[M{\text{ }}\left( {A,{\text{ }}Z} \right)\], \[{M_p}\], and \[{M_n}\] are mass of nucleus, proton and neutron respectively. and Z is mass and atomic number.
Thus, by substituting the value of ∆m in BE, we get
$\begin{gathered}
BE = \{ Z{M_p} + (A - Z){M_n}\} - M(A,Z) \times {c^2} \\
\frac{{BE}}{{{c^2}}} = \{ Z{M_p} + (A - Z){M_n}\} - M(A,Z) \\
M(A,Z) = \{ Z{M_p} + (A - Z){M_n}\} - \frac{{BE}}{{{c^2}}} \\
\end{gathered} $
Hence, the correct option is C
Note The binding energy is expressed in J when mass defect is in kg. And for converting from mass to energy, one needs to multiply \[{c^2}\] to it. To convert from Energy to mass, it has to be divided by \[{c^2}\]. If mass defect is in amu then binding energy is 931 MeV times \[\Delta m\].
In subjects like general relativity and quantum mechanics, the factor \[{c^2}\] that differentiates between mass and energy is ignored; the two quantities are considered equivalent. Just like how the number 1 is ignored after being multiplied to some arbitrary quantity.
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