Nucle

For
$${}^{14}_{7}\mathrm{N},\quad Z=7,\ N=7$$
Mass defect:
$$\Delta M=[7m_p+7m_n]-m(^{14}_{7}\mathrm N)$$
Using the given atomic masses, it is easier to use
$$\Delta M=[7m_H+7m_n]-m(^{14}_{7}\mathrm N)$$
because the electron masses cancel.

$$=7(1.007825)+7(1.008665)-14.00307$$
$$=7(2.01649)-14.00307=14.11543-14.00307=0.11236\,\text{u}$$
Binding energy:
$$E_b=\Delta M\times 931.5$$
$$=0.11236\times 931.5\approx 1.046\times 10^2\,\text{MeV}$$
So the binding energy is about 105 MeV.

Use
$$E_b=\big[Zm_H+(A-Z)m_n-m_{atom}\big]\times 931.5\,\text{MeV}$$

### For $^{56}_{26}\mathrm{Fe}$
$$Z=26,\ A-Z=30$$
$$\Delta M=26(1.007825)+30(1.008665)-55.934939$$
$$=26.20345+30.25995-55.934939=0.528461\,\text{u}$$
$$E_b=0.528461\times 931.5\approx 492\,\text{MeV}$$

### For $^{209}_{83}\mathrm{Bi}$
$$Z=83,\ A-Z=126$$
$$\Delta M=83(1.007825)+126(1.008665)-208.980388$$
$$=83.649475+127.09059-208.980388=1.759677\,\text{u}$$
$$E_b=1.759677\times 931.5\approx 1639\,\text{MeV}$$
So the binding energies are approximately 492 MeV and 1.64\times 10^3 MeV.

For $^{63}_{29}\mathrm{Cu}$, mass number $A=63$. The book gives the approximate nuclear binding energy per nucleon in this region as about 8 MeV.

Energy required for one nucleus:
$$E \approx 63\times 8=504\,\text{MeV}$$

Number of atoms in 3.0 g of copper:
$$\text{moles} = \frac{3.0}{63}=0.04762\,\text{mol}$$
$$N = 0.04762\times 6.023\times 10^{23} \approx 2.87\times 10^{22}$$

Total energy:
$$E_{tot}=2.87\times 10^{22}\times 504\,\text{MeV} \approx 1.45\times 10^{25}\,\text{MeV}$$
Convert to joules using $1\,\text{MeV}=1.6\times 10^{-13}\,\text{J}$:
$$E_{tot}=1.45\times 10^{25}\times 1.6\times 10^{-13} \approx 2.3\times 10^{12}\,\text{J}$$
So the required nuclear energy is of order **$10^{12}$ J; more precisely about $2.3\times 10^{12}$ J**.

If one uses the textbook-style approximate binding energy per nucleon more crudely, the answer is still in the same order, $10^{12}$–$10^{13}$ J. The calculated value above is the best estimate from the given chapter data and standard nuclear binding-energy reasoning.

Nuclear radius varies as
$$R=R_0A^{1/3}$$
So the ratio of radii is
$$\frac{R_{Au}}{R_{Ag}}=\left(\frac{197}{107}\right)^{1/3}$$
$$\frac{197}{107}\approx 1.84$$
$$1.84^{1/3}\approx 1.21$$
Therefore, the radius of gold nucleus is about 1.21 times that of silver nucleus.

The chapter states that the Q-value of a nuclear process is the difference between the initial and final rest masses multiplied by $c^2$:
$$Q = [m_A+m_b-m_C-m_d]c^2$$
It is also equal to the difference between final and initial kinetic energies:
$$Q = \text{final kinetic energy} - \text{initial kinetic energy}$$
If $Q>0$, the reaction is exothermic; if $Q<0$, it is endothermic.