Answer :
Answer:
correct option is C i.e F_{fes} [F] > F_{fes} [He]
Explanation:
Given data:
excited state of helium is given as 159,850 cm^{-1}
In terms of joule = [tex]6.626\times 10^{-34} \times 3\times 10^{10} \times 159,850[/tex]
[tex]= 3.177\times 10^{-18} J[/tex]
First excited state of fluorine is [tex]404 cm^{-1}[/tex]
in terms of joule is [tex]= 6.626\times 10^{-34} \times 3\times 10^{10} \times 404[/tex]
[tex]= 8.030\times 10^{-21} J[/tex]
so, [tex]F_{fes} [He] = e^{- \frac {3.177\times 10^{-18}}{1.38\times 10^{-23}\times 1000}[/tex]
where T =1000 k
[tex]F_{fes} [He] = 0[/tex]
[tex]F_{fes} [He] = e^{- \frac {8.030\times 10^{-21}}{1.38\times 10^{-23}\times 1000}[/tex]
where T = 1000 k
[tex]F_{fes} [F] = 0.55[/tex]
Therefore . [tex]F_{fes} [F] > F_{fes} [He][/tex]
correct option is C