A) \[\Delta {{M}_{A}}+\Delta {{M}_{B}}=\Delta {{M}_{C}}-\Delta E/{{c}^{2}}\]
B) \[\Delta {{M}_{A}}+\Delta {{M}_{B}}=\Delta {{M}_{C}}+\Delta E/{{c}^{2}}\]
C) \[\Delta {{M}_{A}}-\Delta {{M}_{B}}=\Delta {{M}_{C}}-\Delta E/{{c}^{2}}\]
D) \[\Delta {{M}_{A}}-\Delta {{M}_{B}}=\Delta {{M}_{C}}+\Delta E/{{c}^{2}}\]
Correct Answer: A
Solution :
Binding energy of nuclei \[A=\Delta {{M}_{A}}{{C}^{2}}\] Binding energy of nuclei\[B=\Delta {{M}_{B}}{{C}^{2}}\] Binding energy of nuclei\[C=\Delta {{M}_{C}}{{C}^{2}}\] According to question \[A+B\xrightarrow[{}]{{}}C\] So, released energy \[\Delta E=BE\]of C (BE of A and B) \[\Delta {{M}_{C}}{{C}^{2}}-(\Delta {{M}_{A}}{{C}^{2}}+\Delta {{M}_{B}}{{C}^{2}})\] Or \[\frac{\Delta E}{{{c}^{2}}}=\Delta {{M}_{C}}-(\Delta {{M}_{A}}+\Delta {{M}_{B}})\] \[\Delta {{M}_{A}}+\Delta {{M}_{B}}=\Delta {{M}_{C}}-\frac{\Delta E}{{{c}^{2}}}\]
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