Answer
Binding energy per nucleon for $ {^{86}_{36}Kr}, \frac{E_{B}}{A}=8.73\: MeV/nucleon$
Binding energy per nucleon for $ {^{180}_{73}Ta}, \frac{E_{B}}{A}=8.08\: MeV/nucleon$
The results confirm that for A greater than 62, the binding energy per nucleon decreases as A increases.
Work Step by Step
The binding energy per nucleon is the energy required to pull a nucleon from the nucleus of an atom. According to the liquid drop model of an atom, the binding energy is given by
$E_{B} =C_{1}A-C_{2}A^{\frac{2}{3}}-C_{3}\frac{Z(Z-1)}{A^{\frac{1}{3}}}-C_{4}\frac{(A-2Z)^{2}}{A} \pm C_{5}A^{-\frac{4}{3}}$
Where $A$ is the mass number and $Z$ is the atomic number.
$C_{1}, C_{2}, C_{3}, C_{4}$ and $C_{5}$ are constants and has the values
$C_{1}=15.75 \:MeV$
$C_{2}=17.80 \:MeV$
$C_{3}=0.7100 \:MeV$
$C_{4}=23.69 \:MeV$
$C_{5}=39 \:MeV$
The last term in the equation will be positive if both Z and N=A-Z are even and negative if both Z and N are odd, and zero otherwise.
In our problem, we need to first calculate the binding energy of both the given nuclei using the above equation. Then dividing this binding energy by the number of nucleons (mass number) will give us the binding energy per nucleon for the two nuclei.
For $ {^{86}_{36}Kr}$
$A=86$
$Z=36$
$N=50$
here both Z and N are even, thus the last term in the equation will have a positive sign.
Substituting the values for known parameters, we get
$E_{B} =(15.75\times86)-(17.80\times86^{\frac{2}{3}})-(0.7100\times\frac{36(36-1)}{86^{\frac{1}{3}}})-(23.69\times\frac{(86-2\times36)^{2}}{86}) + (39\times86^{-\frac{4}{3}})$
$E_{B} =1354.5-346.81-202.67-53.99+0.10=751.13\:MeV$
Binding energy per nucleon $=\frac{E_{B}}{A}=\frac{751.13}{86}=8.73\: MeV/nucleon$
For $ {^{180}_{73}Ta}$
$A=180$
$Z=73$
$N=107$
here both Z and N are odd, thus the last term in the equation will have a negative sign.
Substituting the values for known parameters, we get
$E_{B} =(15.75\times180)-(17.80\times180^{\frac{2}{3}})-(0.7100\times\frac{73(73-1)}{180^{\frac{1}{3}}})-(23.69\times\frac{(180-2\times73)^{2}}{180}) - (39\times180^{-\frac{4}{3}})$
$E_{B} =2835-567.46-660.93-152.14-0.04=1454.43\:MeV$
Binding energy per nucleon $=\frac{E_{B}}{A}=\frac{1454.43}{180}=8.08\: MeV/nucleon$
By comparing both binding energies per nucleon, it is clear that $ {^{180}_{73}Ta}$ has less binding energy per nucleon than $ {^{86}_{36}Kr}$, which confirms that the binding energy per nucleon decreases as A increases.
