The binding energy per nucleon of $^{209}_{83}Bi$ is . . . . . . MeV. [Take $m(^{209}_{83}Bi) = 208.980388 \text{ u}$, $m_p = 1.007825 \text{ u}$, $m_n = 1.008665 \text{ u}$, $1 \text{ u} = 931 \text{ MeV}/c^2$]

Assuming the experimental mass of $^{12}_{6}C$ as $12 \text{ u}$, the mass defect of $^{12}_{6}C$ atom is . . . . . . $\text{u}$. (Mass of proton $= 1.00727 \text{ u}$, mass of neutron $= 1.00866 \text{ u}$).

The mass defect in a particular nuclear reaction is $0.3 \,g$. The amount of energy liberated in kilowatt-hour $(kWh)$ is: (Velocity of light $c = 3 \times 10^8 \,m/s$)

If $M_{O}$ is the mass of an oxygen isotope ${ }_{8}^{17}O$, $M_{p}$ and $M_{n}$ are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei $_{20}^{41} Ca$ and $_{13}^{27} Al$ from the following data:
$m(_{20}^{40} Ca) = 39.962591 \; u$
$m(_{20}^{41} Ca) = 40.962278 \; u$
$m(_{13}^{26} Al) = 25.986895 \; u$
$m(_{13}^{27} Al) = 26.981541 \; u$
(Given mass of neutron $m_n = 1.008665 \; u$)

$A$ given coin has a mass of $3.0\; g$. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity, assume that the coin is entirely made of $_{29}^{63} Cu$ atoms (of mass $62.92960\; u$).