91Ó°ÊÓ

-$\text{1amu}\text{=}\text{931.5MeV}$

- The atomic mass of${}^{\text{206}}\text{Pbis205.974440amu}$

- Atomic number of$\text{Pb}$is$82$, so it has$82$protons

- The number of nucleons is$206$, therefore,

the number of neutrons is ($206$-$82$) $=124$.

- The mass of neutron is${\text{m}}_{\text{n}}\text{= 1.008665amu}$

- The mass of proton is${\text{m}}_{\text{p}}\text{= 1.007825amu}$

Now, let us calculate the change in mass

$$\begin{gathered} ∆\text{m = 205.974440amu -}\left({\text{82׳¾}}_{\text{p}}{\text{+ 124×m}}_{\text{n}}\right) \\ =\text{205.974440amu -}\left(\text{82×1.007825amu+ 124×1.008665amu}\right) \\ =\text{- 1.74167amu} \end{gathered}$$

(a)

The binding energy per nucleon, in $\text{MeV}$is

$\text{BE per nucleon =}\frac{{\displaystyle \text{1.74167²¹³¾³Ü×}\frac{\text{931.5MeV}}{\text{1amu}}}}{{\displaystyle \text{206nucleons}}}\text{= 7.926MeV/ nucleon}$

(b)

The binding energy per atom is

$\text{BE per atom =}\text{7.926MeV/ nucleon×206 nucleons/atom =}\text{1632.76MeV/ atom}$

(c)

The binding energy per mole in

$$\begin{gathered} \text{BE per mol in kJ =}{\text{BE per atom×6.022×10}}^{\text{23}}\text{atoms /mol} \\ \text{=}\text{1632.76MeV/ atom}{×6.022×10}^{\text{23}}\text{atoms /mol} \\ ={\text{9.83×10}}^{\text{26}}\text{²Ñ±ð³Õ/³¾´Ç±ô×}\frac{{\displaystyle {\text{1.602×10}}^{\text{- 13}}\text{J}}}{{\displaystyle \text{1MeV}}} \\ ={\text{1.575×10}}^{\text{14}}\text{´³/³¾´Ç±ô×}\frac{\text{1kJ}}{\text{1000J}} \\ ={\text{1.575×10}}^{\text{14}}\mathrm{kJ}/\mathrm{mol} \end{gathered}$$