Find the binding energy and binding energy per nucleon of the nucleus ${}_{28}{}^{62}\mathrm{Ni}$. The atomic mass of nickel is $61.9283 \mathrm{u}$.

A fission reaction involving uranium is:

${}_{92}{}^{235}\mathrm{U}+{}_0{}^1\mathrm{n}\to {}_{40}{}^{98}\mathrm{Zr}+{}_{52}{}^{135}\mathrm{Te}+3{}_0{}^1\mathrm{n}$

Calculate the energy released.(in $\mathrm{MeV}$) (Atomic masses: $\mathrm{U}=235.043922 \mathrm{u}$; $\mathrm{Zr}=97.91276 \mathrm{u}$; $\mathrm{Te}=134.9165 \mathrm{u}$)

Calculate the energy released (in $\mathrm{MeV}$) in the fusion reaction:

${}_1{}^{2}H+{}_1{}^{3}H\to {}_2{}^{4}He+{}_0{}^1\mathrm{n}$

(Atomic masses:${}_1{}^{2}H=2.014102 \mathrm{u}$, ${}_1{}^{3}H=3.016049 \mathrm{u}$, ${}_2{}^{4}He=4.002603 \mathrm{u}$)

In the first nuclear reaction in a particle accelerator, hydrogen nuclei were accelerated and then allowed to hit nuclei of lithium according to the reaction:

${}_1{}^1\mathrm{H}+{}_3{}^7\mathrm{Li}\to {}_2{}^4\mathrm{He}+{}_2{}^4\mathrm{He}$

Calculate the energy released. (The atomic mass of lithium is $7.016 \mathrm{u}$, the atomic mass of helium is $4.002603 \mathrm{u}$ and atomic mass of ${}_1{}^1\mathrm{H} \mathrm{is} 1.007825 \mathrm{u}$)

Show that an alternative formula for the mass defect is $\delta =Z{M}_H+\left(A-Z\right)m_n-M_{atom}$ where $M_H$ is the mass of a hydrogen atom and $m_n$ is the mass of a neutron.

Consider the nuclear fusion reaction involving the deuterium $\left({}_1{}^2\mathrm{D}\right)$ and tritium $\left({}_1{}^3\mathrm{T}\right)$ isotopes f hydrogen:

${}_1{}^2\mathrm{D}+{}_1{}^3\mathrm{T}\to {}_2{}^4\mathrm{He}+{}_0{}^1\mathrm{n}$

The energy released $\mathrm{Q}$  can be calculated in a usual way, using the masses of particles involved, from the following expression:

$Q=\left(M_D+M_T-M_{He}-m_n\right) c^2$

Show that the expression for $Q$ can be rewritten as:

$Q=E_{He}-\left(E_D+E_T\right)$

where $E_{He}, E_D \mathrm{and} {\mathrm{E}}_{\mathrm{T}}$ are the binding energies of helium, deuterium, and tritium, respectively.

The fission reaction of uranium:

${}_{92}{}^{235}\mathrm{U}+{}_0{}^1\mathrm{n}\to {}_{40}{}^{98}\mathrm{Zr}+{}_{52}{}^{135}\mathrm{Te}+3{}_0{}^1\mathrm{n}$

The energy released, $Q$, may be calculated from:

$Q=\left(M_U-M_{Zr}-M_{Te}-2m_n\right)c^2$

Show that the expression for $Q_2$ can be rewritten as:

$Q=\left(E_{\mathrm{Zr}}+E_{\mathrm{Te}}\right)-E_{\mathrm{U}}$

where $E_{\mathrm{Zr}}, E_{\mathrm{Te}} \mathrm{and} E_{\mathrm{U}}$ are biding energies of zirconium, tellurium, and uranium respectively.

Explain using the binding energy curve, why energy is released in fusion and fission reactions.