A gold nucleus 19779Au absorbs a neutron of nearly zero kine

A gold nucleus (¹⁹⁷/₇₉)Au absorbs a neutron of nearly zero kinetic energy, and emits an electron.

A: What is produced?

B: How much energy is released in this reaction, in eV?

Solution

¹⁹⁷₇₉ Au + ¹₀ n ---> ⁰_-1 e + ¹⁹⁸₈₀ X

Where X = mercury     Since atomic number 80 is mercury

              = Hg

(B).Mass of ¹⁹⁷₇₉ Au , m = 196.9665687 u

mass of neutron , m \' = 1.008665 u

Mass of ¹⁹⁸₈₀ Hg is M = 197.9667690 u

Mass of electron M \' = 0.000548579909 u

Mass defect dm = m+m\' - M - M\'

                        = 9.013279 x10 ^-3 u

We know 1 u = 931.5 MeV

So, energy released E = dm x 931.5 MeV

                                 = 9.013279 x10 ^-3 x931.5 MeV

                                 = 8.395 MeV

                                = 8.395 x10 ⁶ eV