A) \[28.4\text{ }MeV\]
B) \[0.061\text{ }u\]
C) \[0.0305\text{ }J\]
D) \[0.0305\text{ }erg\]
Correct Answer: A
Solution :
| Key Idea: The binding energy of a nucleus, |
| \[\Delta E=\] (mass defect) \[\times 931\] |
| \[_{2}H{{e}^{4}}\] contains 2 neutrons and 2 protons |
| Mases of 2 protons \[=2\times 1.0073\] |
| \[=2.0146\text{ }u\] |
| Mass of 2 neutrons \[=2\times 1.0087\] |
| \[=2.0174\text{ }u\] |
| Total mass of 2 protons and 2 neutrons |
| \[=\left( 2.0146+2.0174 \right)U=4.032\text{ }u\] |
| Mass of helium nucleus \[=4.0015\text{ }u\] |
| Thus, mass defect is lacking of mass in forming the helium nucleus from 2 protons and 2 neutrons. |
| \[\therefore \] \[\Delta m=\] mass defect |
| \[=\left( 4.032-4.0015 \right)\text{ }u\] |
| \[=0.0305\text{ }u\] |
| \[=0.0305\text{ }amu\] |
| but 1 amu \[=931\text{ }MeV\] |
| Hence, binding energy |
| \[\Delta E=(\Delta m)\times 931\] |
| \[=0.0305\times 931=28.4\,MeV\] |
| Given, mass of helium nucleus \[\approx ~4.0015\text{ }u\] |
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