Respuesta :
Explanation:
It is given that r = 0.283 nm. As 1 nm = [tex]10^{-9} m[/tex].
Hence, 0.283 nm = [tex]0.283 \times 10^{-9} m[/tex]
- Formula for coulombic energy is as follows.
[tex]U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}[/tex]
where, e = [tex]1.6 \times 10^{-19}[/tex] C
[tex]\epsilon_{o}[/tex] = [tex]8.85 \times 10^{-12}[/tex]
[tex]U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}[/tex]
= [tex]1.423 \times 10^{-18} J[/tex]
- As 1 eV = [tex]1.6 \times 10^{-19} J[/tex]
So, 1 J = [tex]\frac{1 eV}{1.6 \times 10^{-19}}[/tex]
Hence, U = [tex]\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}[/tex]
= 8.9 eV
- Also, 1 J = [tex]\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}[/tex]
= [tex]1.67 \times 10^{-27}[/tex] kJ/mol
Therefore, U = [tex]1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}[/tex] kJ/mol
= [tex]2.37 \times 10^{-45} kJ/mol[/tex]
