Find the binding energy of a nucleus consisting of equal numbers of protons and neutrons, and having the radius one and a half times smaller than that of ${\mathrm{Al}}^{27}$ nucleus.

Mass of the nucleon is $1.66\times {10}^{-27} \mathrm{kg}$, $R_0=1.3\times {10}^{-15} \mathrm{m}$.

 $m_{\mathrm{H}}=1.007825$ for hydrogen, $m_{\mathrm{N}}=1.008665$ for neutron, $M=8.00531$ for $\mathrm{Be}$. 

Making use of the tables of atomic masses, find 
$\left(\mathrm{a}\right)$ the mean binding energy per one nucleon in ${\mathrm{O}}^{16}$ $(M=15.9994 \mathrm{amu})$ nucleus, 
$\left(\mathrm{b}\right)$ the binding energy of a neutron and an alpha particle in a ${\mathrm{B}}^{11}$ $(M=11.00930 \mathrm{amu})$ nucleus, 
$\left(\mathrm{c}\right)$ the energy required for separation of an ${\mathrm{O}}^{16}$ nucleus into four identical particles.

$$\begin{gathered} m_{\mathrm{proton}}=1.007823 \mathrm{amu} \\ {m}_{\mathrm{neutron}}=1.008665 \mathrm{amu} \\ {m}_{\mathrm{\alpha }-\mathrm{particle}}=4.00260 \mathrm{amu} \end{gathered}$$

Find the difference in binding energies of a neutron and a proton in a ${\mathrm{B}}^{11}$ nucleus. Explain why there is the difference.

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931.5 \mathrm{MeV}$ of energy.

$m_{\mathrm{n}}=1.008665 \mathrm{amu},$ $m_{\mathrm{p}}=1.007276 \mathrm{amu},$ $m_{{\mathrm{B}}_{10}}=10.01294 \mathrm{amu},$ $m_{{\mathrm{Be}}_{10}}=10.01354 \mathrm{amu}$. 

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931 \mathrm{MeV}$ of energy.

Assuming that the splitting of a ${\mathrm{U}}^{235}$ nucleus liberates the energy of $200 \mathrm{MeV}$, find,

(a) the energy liberated in the fission of one kilogram of ${\mathrm{U}}^{235}$ isotope and the mass of coal with calorific value of $30 \mathrm{kJ} {\mathrm{g}}^{-1}$ which is equivalent to that for one $\mathrm{kg}$ of ${\mathrm{U}}^{235}$.
(b) the mass of ${\mathrm{U}}^{295}$ isotope split during the explosion of the atomic bomb with $30 \mathrm{kt}$ trotyl equivalent if the calorific value of trotyl is $4.1 \mathrm{kJ} {\mathrm{g}}^{-1}$.

Avogadro's number$=6.023\times {10}^{23}$

Avogadro number$=6.023\times {10}^{23}$, $1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931.5 \mathrm{MeV}$ of energy.

Excess of the mass of atoms $∆{\mathrm{Li}}^6=0.01513 \mathrm{amu}, ∆{\mathrm{H}}^2=0.01410 \mathrm{amu} and ∆{\mathrm{He}}^4=0.00260 \mathrm{amu}$ and of neutron and proton are$∆{\mathrm{m}}_{\mathrm{p}}=0.00783 \mathrm{amu} \mathrm{and} ∆{\mathrm{m}}_{\mathrm{n}}=0.00867 \mathrm{amu}$

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931.5 \mathrm{MeV}$ of energy.