A semiempirical formula for the binding energy of a nucleus of atomic number Z and mass number A is E, = C1A – C2A²/3 – C3 Z(Z – 1) (A – 2Z)? - C4 +C,A-4/3, A1/3 A where C = 14.1 MeV, C2 = 13.0 MeV, C3 = 0.595 MeV, C4 = 19.0 MeV, and C5 = 33.5 MeV. Find the most stable isobar for A = 25.

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A semiempirical formula for the binding energy of a nucleus of atomic number Z and mass number A is
Z(Z – 1)
A!/3
(A – 2Z)²
- C4
E, = C1A – C2A²/3 – C3
±C;A¬4/3,
where C = 14.1 MeV, C2 = 13.0 MeV, C3 = 0.595 MeV, C4 = 19.0 MeV, and C5 = 33.5 MeV. Find the most
stable isobar for A = 25.
Transcribed Image Text:A semiempirical formula for the binding energy of a nucleus of atomic number Z and mass number A is Z(Z – 1) A!/3 (A – 2Z)² - C4 E, = C1A – C2A²/3 – C3 ±C;A¬4/3, where C = 14.1 MeV, C2 = 13.0 MeV, C3 = 0.595 MeV, C4 = 19.0 MeV, and C5 = 33.5 MeV. Find the most stable isobar for A = 25.
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