Question

For each of the following nuclei, determine the binding energy per nucleon (in MeV). (For all masses, keep six places beyond the decimal point when performing your calculations. Then round your final answer to at least three significant figures.)(a) 21 MeV (b) 4He MeV (c) 180 MeV (d) 23Na MeV

Answer 1

Answer: (a) BE = 1.112 MeV

               (b) BE = 7.074 MeV

               (c) BE = 7.767 MeV

               (d) BE = 8.112 MeV

Explanation: Binding energy per nucleon is the average energy necessary to remove a proton or a neutron from the nucleus of an atom. It is mathematically defined as:

$BE = \frac{\Delta m.c^{2}}{A}$

Where

Δm is a difference in mass known as mass defect

A is atomic mass of an atom.

Mass Defect is determined by:

$\Delta m =Zm_{p}+(A-Z)m_{n} - m_{nuc}$

where:

Z is atomic number

$m_{p}$ is mass of proton

$m_{n}$ is mass of neutron

$m_{nuc}$ is mass of the nucleus

Mass of proton is 1.007825u.

Mass of neutron is 1.008665u.

The unit u is equal to 931.5MeV/c².

(a) 2H(deuterion): Given: Z = 1; A = 2; $m_{nuc}$ = 2.014102u

$\Delta m =1(1.007825)+1(1.008665) -2.014102$

$\Delta m =0.002388u$

$BE = \frac{0.002388.c^{2}}{2}.931.5\frac{MeV}{c^{2}}$

BE = 1.112MeV

(b) 4He (Helium): Given: Z = 2; A = 4; $m_{nuc}$ = 4.002603

$\Delta m =2(1.007825)+2(1.008665) -4.002603$

$\Delta m =0.030377u$

$BE = \frac{0.030377.c^{2}}{4}.931.5\frac{MeV}{c^{2}}$

BE = 7.074MeV

(c) 18O (Oxygen): Given: Z = 8; A = 18; $m_{nuc}$ = 17.999160

$\Delta m =8(1.007825)+10(1.008665) -17.999160$

$\Delta m =0.15009u$

$BE = \frac{0.15009.c^{2}}{18}.931.5\frac{MeV}{c^{2}}$

BE = 7.767MeV

(d) 23Na (Sodium): Given: Z = 11; A = 23; $m_{nuc}$ = 22.989767

$\Delta m =11(1.007825)+12(1.008665) -22.989767$

$\Delta m =0.200288u$

$BE = \frac{0.200288.c^{2}}{23}.931.5\frac{MeV}{c^{2}}$

BE = 8.112MeV