The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u = atomic mass unit) The binding energy of 2H^4 is (mass of helium nucleus = 4.0015 u)
28. 4 MeV
0.061 u
0.0305 J
0.0305 erg
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18
Q]____
=> a) 2He^4
2He^4 contains 2 neutrons and 2 protons
So, mass of 2 protons = 2× 1.0073 =2.0146 u
So, mass of 2 neutrons = 2× 1.0087
=2.0174 u
Total mass of two protons and two neutrons
= (2.0146 + 2.0174 ) u = 4.032 u
Mass of helium nucleus = 4.032 u
Thus, mass defect is lacking of mass in forming the helium nucleus from 2 protons and 2 neutrons.
∴ ∆m = mass defect = (4.032 - 4.0015)u
= 0.0305 u
As we know that, 1u = 931 MeV
Hence, binding energy
∆E = ( ∆m ) × 931
= 0.0305 × 931
= 28.4 MeV
∴ The binding Energy ∆E = 28.4 MeV !
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1
Answer:
♣️The binding energy ∆E is 28.4 MeV..♣️
Explanation:
♥️Hope it will be helpful ☺️ thank you!!♥️
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