What is the disintegration energy (in MeV) associated with this spontaneous decay? ${ }_{60}^{144} \mathrm{Nd} \rightarrow{ }_{58}^{140} \mathrm{Ce}+{ }_{2}^{4} \mathrm{He}$ mNd = 143.910083 u mCe = 139.905434 u
MHe = 4.002603 u
1 u = 1.66 * 10-27 kg
A)1.54
B)2.37
C)1.90
D)4.13
E)8.21

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The Answer of What is the disintegration energy (in MeV) associated with this spontaneous decay? ${ }_{60}^{144} \mathrm{Nd} \rightarrow{ }_{58}^{140} \mathrm{Ce}+{ }_{2}^{4} \mathrm{He}$ mNd = 143.910083 u mCe = 139.905434 u
MHe = 4.002603 u
1 u = 1.66 * 10-27 kg
A)1.54
B)2.37
C)1.90
D)4.13
E)8.21

What Is the Disintegration Energy (In MeV) Associated with This 60144Nd→58140Ce+24He{ }_{60}^{144} \mathrm{Nd} \rightarrow{ }_{58}^{140} \mathrm{Ce}+{ }_{2}^{4} \mathrm{He}60144​Nd→58140​Ce+24​He

What Is the Disintegration Energy (In MeV) Associated with This ${ }_{60}^{144} \mathrm{Nd} \rightarrow{ }_{58}^{140} \mathrm{Ce}+{ }_{2}^{4} \mathrm{He}$