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 defect) \[\times \] 931 \[_{2}H{{e}^{4}}\] contains 2 neutrons and 2 protons Mases 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 = 4.032 u 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 defect = (4.032 - 4.0015) u = 0.0305 u = 0.0305 amu but 1 amu = 931 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 u
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