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Answer:
Explanation:
From the given information:
The energy of photons can be determined by using the formula:
[tex]E = \dfrac{hc}{\lambda}[/tex]
where;
planck's constant (h) = [tex]6.63 \times 10^ {-34}[/tex]
speed oflight (c) = [tex]3.0 \times 10^8 \ m/s[/tex]
wavelength λ = 58.4 nm
[tex]E = \dfrac{6.63 \times 10^{-34} \ J.s \times 3.0 \times 10^8 \ m/s}{58.4 \times 10^{-9 } \ m}[/tex]
[tex]E =0.34 \times 10^{-17} \ J[/tex]
[tex]E = 3.40 \times 10^{-18 } \ J[/tex]
To convert the energy of photon to (eV), we have:
[tex]1 eV = 1.602 \times 10^{-19} \ J[/tex]
Hence
[tex]3.40 \times 10^{-18 } \ J = \dfrac{1 eV}{1.602 \times 10^{-19 } \ J }\times 3.40 \times 10^{-18 } \ J[/tex]
[tex]E = 2.12 \times 10 \ eV[/tex]
E = 21.2 eV
b)
The equation that illustrates the process relating to the first ionization is:
[tex]Hg_{(g)} \to Hg^+ _{(g)} + e^-[/tex]
c)
The 1st ionization energy (I.E) of Hg can be calculated as follows:
Recall that:
I.E = Initial energy - Kinetic Energy
I₁ (eV) = 21.2 eV - 10.75 eV
I₁ (eV) = 10.45 eV
Since ;
[tex]1 eV = 1.602 \times 10^{-19} \ J[/tex]
∴
[tex]10.45 \ eV = \dfrac{1.602 \times 10^{-19 } \ J }{ 1 \ eV}\times 10.45 \ eV[/tex]
Hence; the 1st ionization energy of Hg atom = [tex]1.67 \times 10^{-18} \ J[/tex]
[tex]1.67 \times 10^{-21} \ kJ[/tex]
Finally;
[tex]I_1 \ of \ the \ Hg (kJ/mol) = \dfrac{1.67 \times 10^{-21 \ kJ} \times 6.02 \times 10^{23} \ Hg \ atom }{1 \ Kg \ atom }[/tex]
[tex]\mathbf{= 1.005 \times 10^3 \ kJ/mol}[/tex]