Calculate the binding energy, in MeV, of a nucleus of $^{59}_{27}$Co - AQA - A-Level Physics - Question 5 - 2017 - Paper 2

To find the binding energy, we use the mass defect formula:

$$ext{Binding Energy} = ( ext{Mass of protons} + ext{Mass of neutrons} - ext{mass of nucleus}) c^2$$

1. Calculate the number of protons and neutrons in $^{59}_{27}$Co:

   • Number of protons (Z) = 27
   • Number of neutrons (N) = 59 - 27 = 32

2. Calculate the mass of protons and neutrons:

   • Mass of protons = $27 imes 1.00728$ u = 27.19656 u
   • Mass of neutrons = $32 imes 1.00866$ u = 32.27712 u

3. Sum the masses of protons and neutrons:

   • Total mass = $27.19656$ u + $32.27712$ u = 59.47368 u

4. Calculate the mass defect:

   • Mass defect = $59.47368$ u - $58.93320$ u = $0.54048$ u

5. Convert the mass defect to energy:

   • Using Einstein's equation $E = mc^2$, where $1 ext{ u} = 931.5 ext{ MeV}$:
   • Binding Energy = $0.54048 ext{ u} imes 931.5 ext{ MeV/u} = 503.83 ext{ MeV}$

Thus, the binding energy of $^{59}_{27}$Co is approximately 503.83 MeV.