find the energy in joules

Question

find the binding energy in joules of a cobalt atom with 27 protons, 32 neutrons and a mass of 97.958205x10^-27kg

Answer 1

The binding energy of an atom refers to the binding energy of its nuceleus, which is defined by the Einstein's mass-energy relation

 

                                           ΔE = (Δ m)c^{2                  (1)}

 

where Δ E is the binding energy, Δ m - mass defect (difference between combined mass of free protons and neutrons, and the atomic nucleus, c - speed of light. Let's calculate the mass of free protons and neutrons since the mass of the nucleus is given to you - 97.958205 x 10^-27kg. 

 

The mass m_p of one proton is approximately 1.66 x 10-27 kg, and of the neutron (m_n )is

 

1.6612 x 10-27 kg. Thus, mass of all free protons and neutrons together in the atom of Cobalt is

 

27x m_p + 32 m_n  = 44.82 x 10^-27 kg + 53.16 x 10^-27 kg = 97.98 x 10^-27kg.

 

Thus, the mass deffect is 

 

     Δm = (97.98 - 97.958205)x 10^-27 kg = 0.032 x 10^-27 kg

 

Now, substituting this number into (1) with the speed of light c = 3 x 10⁸ m/s we can obtain 

 

                Δ E = 2.88 X 10^-12 J = 18 Mev (megaelectronvolt)

 

(1 ev = 1.6 x 10-19 J).