Obtain the binding energy of the nuclei $_{26}^{56} F e$ and $_{83}^{209} B i$ in units of MeV from the following data:
(a) $m (_{26}^{56} F e) = 55.934939 u$
(b) $m (_{83}^{209} B i) = 208.980388 u$

Question

Obtain the binding energy of the nuclei 2656​Fe and 83209​Bi in units of MeV from the following data:(a) m(2656​Fe)=55.934939u(b) m(83209​Bi)=208.980388u

Filo · Accepted Answer

(a) 2656​Fe contains 26 protons and 30 neutrons.Mass of 26 protons=26×1.007825u=26.26345uMass of 30 neutrons=30×1.008665u=30.25995uMass of 2656​Fe nucleus=55.934939uThus, mass defect Δm=26.26345u+30.25995u−55.934939u=0.528461u.Thus, binding energy of 2656​Fe nucleus is 0.528461×931.5MeV=492.26MeV.Binding energy per nucleon=56492.26MeV​=8.790MeV(b) 83209​Bi contains 83 protons and 126 neutrons.Mass of 83 protons=83×1.007825u=83.649475uMass of 126 neutrons=126×1.008665u=127.09170uMass of 83209​Bi nucleus=208.986388uThus, mass defect Δm=83.649475u+127.09170u−208.986388u=1.760872u.Thu,s binding energy of 83209​Bi nucleus is 1.760872×931.5MeV=1640.26MeV.Binding energy per nucleon=2091640.26MeV​=7.848MeV

Question Text	Obtain the binding energy of the nuclei $_{26}^{56} F e$ and $_{83}^{209} B i$ in units of MeV from the following data: (a) $m (_{26}^{56} F e) = 55.934939 u$ (b) $m (_{83}^{209} B i) = 208.980388 u$
Answer Type	Text solution:1
Upvotes	150

Obtain the binding energy of the nuclei $_{26}^{56} F e$ and $_{83}^{209} B i$ in units of MeV from the following data:
(a) $m (_{26}^{56} F e) = 55.934939 u$
(b) $m (_{83}^{209} B i) = 208.980388 u$

Text solutionVerified

Practice questions from similar books

Obtain the binding energy of the nuclei 2656​Fe and 83209​Bi in units of MeV from the following data:(a) m(2656​Fe)=55.934939u(b) m(83209​Bi)=208.980388u

Text solutionVerified

Practice questions from similar books

Obtain the binding energy of the nuclei $_{26}^{56} F e$ and $_{83}^{209} B i$ in units of MeV from the following data:
(a) $m (_{26}^{56} F e) = 55.934939 u$
(b) $m (_{83}^{209} B i) = 208.980388 u$