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