For atomic masses, use $\Delta m=Zm_H+(A-Z)m_n-M_{\text{atom}}$, since electron masses cancel. For $^{14}_{7}\text{N}$, $Z=7$ and $A-Z=7$. Thus $\Delta m=7(1.007825)+7(1.008665)-14.00307=0.11236\,\text{u}$. Binding energy $E_b=\Delta m(931.5\,\text{MeV})=0.11236(931.5)=104.7\,\text{MeV}$.
$1.05\times10^2\,\text{MeV}$, approximately $104.7\,\text{MeV}$.
For $^{56}_{26}\text{Fe}$, $Z=26$ and $N=30$. Thus $\Delta m=26(1.007825)+30(1.008665)-55.934939=0.528461\,\text{u}$, so $E_b=0.528461(931.5)=492\,\text{MeV}$. For $^{209}_{83}\text{Bi}$, $Z=83$ and $N=126$. Thus $\Delta m=83(1.007825)+126(1.008665)-208.980388=1.760877\,\text{u}$, so $E_b=1.760877(931.5)=1.64\times10^3\,\text{MeV}$.
$E_b(^{56}_{26}\text{Fe})\approx492\,\text{MeV}$ and $E_b(^{209}_{83}\text{Bi})\approx1.64\times10^3\,\text{MeV}$.
For $^{63}_{29}\text{Cu}$, $Z=29$ and $N=34$. Its mass defect is $\Delta m=29(1.007825)+34(1.008665)-62.92960=0.591935\,\text{u}$. Binding energy per nucleus is $0.591935(931.5)=551\,\text{MeV}$. The number of copper atoms in $3.0\,\text{g}$ is $(3.0/62.92960)(6.023\times10^{23})=2.87\times10^{22}$. Total energy required is $(2.87\times10^{22})(551)=1.6\times10^{25}\,\text{MeV}=2.5\times10^{12}\,\text{J}$.
$1.6\times10^{25}\,\text{MeV}$, or about $2.5\times10^{12}\,\text{J}$.
Nuclear radius follows $R=R_0A^{1/3}$. Therefore $R_{\text{Au}}/R_{\text{Ag}}=(197/107)^{1/3}=1.23$.
$R_{\text{Au}}/R_{\text{Ag}}\approx1.23$.
For (i), using atomic masses, $Q=[m(^1\text{H})+m(^3\text{H})-2m(^2\text{H})]c^2=[1.007825+3.016049-2(2.014102)](931.5)=-4.03\,\text{MeV}$, so the reaction is endothermic. For (ii), $Q=[2m(^{12}\text{C})-m(^{20}\text{Ne})-m(^4\text{He})]c^2=[24.000000-19.992439-4.002603](931.5)=+4.62\,\text{MeV}$, so the reaction is exothermic.
(i) $Q=-4.03\,\text{MeV}$, endothermic. (ii) $Q=+4.62\,\text{MeV}$, exothermic.
For $^{56}_{26}\text{Fe}\to2\,^{28}_{13}\text{Al}$, $Q=[m(^{56}\text{Fe})-2m(^{28}\text{Al})]c^2=[55.93494-2(27.98191)](931.5)=-26.9\,\text{MeV}$. The negative Q-value means the fission is endothermic and is not energetically possible without external energy.
No. $Q=-26.9\,\text{MeV}$, so energy must be supplied.
The number of atoms in $1\,\text{kg}$ of pure $^{239}\text{Pu}$ is $N=(1000/239)(6.023\times10^{23})=2.52\times10^{24}$. With $180\,\text{MeV}$ released per fission, the total energy is $E=N(180)=4.5\times10^{26}\,\text{MeV}$.
$4.5\times10^{26}\,\text{MeV}$.
Each fusion reaction consumes two deuterium nuclei and releases $3.27\,\text{MeV}$. In $2.0\,\text{kg}$ of deuterium, the number of deuterium atoms is approximately $(2000/2)(6.023\times10^{23})=6.02\times10^{26}$. Thus the number of reactions is $3.01\times10^{26}$. The energy released is $(3.01\times10^{26})(3.27)=9.85\times10^{26}\,\text{MeV}=1.58\times10^{14}\,\text{J}$. A $100\,\text{W}$ lamp uses $100\,\text{J s}^{-1}$, so the time is $1.58\times10^{14}/100=1.58\times10^{12}\,\text{s}$, about $5.0\times10^4$ years.
About $1.6\times10^{12}\,\text{s}$, or $5.0\times10^4$ years.
Each deuteron has charge $+e$. When two deuterons of radius $2.0\,\text{fm}$ just touch, their centre-to-centre separation is $r=4.0\,\text{fm}=4.0\times10^{-15}\,\text{m}$. The Coulomb potential energy is $U=(1/4\pi\epsilon_0)e^2/r=9.0\times10^9(1.6\times10^{-19})^2/(4.0\times10^{-15})=5.8\times10^{-14}\,\text{J}$. In MeV, this is $(5.8\times10^{-14})/(1.6\times10^{-13})=0.36\,\text{MeV}$.
$5.8\times10^{-14}\,\text{J}$, or about $0.36\,\text{MeV}$.
The nuclear radius is $R=R_0A^{1/3}$, so the nuclear volume is $V=\frac43\pi R^3=\frac43\pi R_0^3A$. The nuclear mass is approximately $M=Am_N$, where $m_N$ is the average nucleon mass. Therefore $\rho=M/V=Am_N/(\frac43\pi R_0^3A)=3m_N/(4\pi R_0^3)$, which contains no $A$. Hence nuclear matter density is nearly constant for all nuclei.
Since nuclear mass is proportional to $A$ and nuclear volume is also proportional to $A$, nuclear density is independent of $A$.