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Compute the binding energy of ""_(2)^(4)"He" nucleus using the following data: Atomic mass of Helium atom, M_(A) (He) = 4.00260 u and that of hydrogen atom, m_(H) = 1.00785 u. |
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Answer» Solution :BINDING ENERGY BE = `[Zm_(H) + Nm_(N) - M_(A)]c^(2)` For helium nucleus, `Z = 2, N = A -Z = 4 - 2 = 2` Mass defect `DELTA m = [(2 xx 1.00785u )] + (2 xx 1.008665u ) - 4.00260 u] Delta m = 0.03038 u` `B.E = 0.03038u xx c^(2)` `B.E = 0.03038 xx 931 MeV = 28 MeV` `[therefore 1 uc^(2) = 931 MeV]` The binding energy of the `""_(2)^(4)"He"` nucleus is 28 MeV. |
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