What is the binding energy in kj/mol for 249/12(on bottom) Mg (24.97925 g/mol)?.

Question

What is the binding energy in kj/mol for 249/12(onbottom) Mg (24.97925 g/mol)?....

( 1 H   = 1.00783g/mol……0 n= 1.00867 g/mol……..)4912M

The mass of the   Mg is 24.97925g/mol         

                                               Mg has 12 protons +12 nuetrons

                                              Mass = 12*1.00783 +12*1.00867

                                                                 = 12.09396 + 12.10404

                                                                   = 24.198 g/mol

                                                              = (24.97925- 24.198)

                                                                  = 0.78125 g /mol

               

              Now the energy is calculated by Einstein s equation.

      

                                                        E =m*c2

                                                           = 0.78125 g/mol*(2.9979*108 ) m/s)2 *10-3

Explanation / Answer

                                                          mc2   = 7.02*1013 kJ   for given reaction.         But binding energyfor the nuecleons in the Mg   is                                                                    B.E   =   mc2 / Total nucleons= (P + N) in Mg                                                                    = 7.02*1013 /12+12                                                                    = 7.02*1013 / 24                                                B.E          =0.2925*1013 kJ

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