91Ó°ÊÓ

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

-The atomic mass of${}^{\text{131}}\text{Iis130.906114amu}$

- Atomic number of$\text{Iis53}$, so it has$\text{53}$protons

- The number of nucleons is$131$, therefore,

the number of neutrons is ($131$-$53$)$=78$

- 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

$$\begin{gathered} ∆\text{m =}\text{130.906114amu -}\left({\text{53׳¾}}_{\text{p}}{\text{+ 78×m}}_{\text{u}}\right) \\ =\text{130.906114amu -}\left(\text{53×1.007825amu+ 78×1.008665amu}\right) \\ =\text{- 1.184481amu} \end{gathered}$$

(a)

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

$\text{BE per nucleon =}\frac{{\displaystyle \text{1.184481²¹³¾³Ü×}\frac{\text{931.5MeV}}{\text{1amu}}}}{{\displaystyle \text{131nucleons}}}\text{=}\text{8.422MeV/ nucleon}$

(b)

The binding energy per atom is

$\begin{array}{rcl}\text{BE per atom}& =& \text{8.422MeV/ nucleon×131nucleons/atom}\\ & =& \text{1103.282MeV/ atom}\end{array}$

(c)

The binding energy per mole in $\text{kJ}$

$\begin{array}{rcl}\text{BE per mol in kJ}& =& {\text{BE per atom×6.022×10}}^{\text{23}}\text{atoms /mol}\\ & =& \text{1103.282MeV/ atom}{×6.022×10}^{\text{23}}\text{atoms /mol}\\ & =& {\text{6.644×10}}^{\text{26}}\text{²Ñ±ð³Õ/³¾´Ç±ô×}\frac{{\displaystyle {\text{1.602×10}}^{\text{- 13}}\text{J}}}{{\displaystyle \text{1MeV}}}\\ & =& {\text{1.064×10}}^{\text{14}}\text{´³/³¾´Ç±ô×}\frac{\text{1kJ}}{\text{1000J}}\\ & =& {\text{1.064×10}}^{\text{11}}\mathrm{kJ}/\mathrm{mol}\end{array}$