Respuesta :
For Nitrogen:
wavelength = NA x h C / bond energy
= 6.022 x 10²³ x 6.626 x 10⁻³⁴ x 3 x 10⁸ / (945 x 10³)
= 1.27 x 10⁻⁷ m
For Oxygen::
wavelength = NA x h C / bond energy
= 6.022 x 10²³ x 6.626 x 10⁻³⁴ x 3 x 10⁸ / (498 x 10³)
= 2.40 x 10⁻⁷ m
For Fluorine:
wavelength = NA x h C / bond energy
= 6.022 x 10²³ x 6.626 x 10⁻³⁴ x 3 x 10⁸ / (159 x 10³)
= 7.52 x 10⁻⁷ m
(NA is Avogadro's number, h is Planck's constant and C is speed of light )
wavelength = NA x h C / bond energy
= 6.022 x 10²³ x 6.626 x 10⁻³⁴ x 3 x 10⁸ / (945 x 10³)
= 1.27 x 10⁻⁷ m
For Oxygen::
wavelength = NA x h C / bond energy
= 6.022 x 10²³ x 6.626 x 10⁻³⁴ x 3 x 10⁸ / (498 x 10³)
= 2.40 x 10⁻⁷ m
For Fluorine:
wavelength = NA x h C / bond energy
= 6.022 x 10²³ x 6.626 x 10⁻³⁴ x 3 x 10⁸ / (159 x 10³)
= 7.52 x 10⁻⁷ m
(NA is Avogadro's number, h is Planck's constant and C is speed of light )
Explanation:
Longest wavelengths of light that can cleave the bonds in elemental nitrogen
Energy to cleave 1 mol N-N bond = 945 kJ/mol = 945000 J/mol
1 mol= [tex]6.022\times 10^{23}[/tex]
Energy to break 1 N-N bond = [tex]\frac{945000 J/mol}{6.022\times 10^{23} mol^{-1}}=1.56\times 10^{-18} J[/tex]
[tex]E=\frac{hc}{\lambda }[/tex]
[tex]\lambda =\frac{6.626\times 10^{-34}J s\times 3\times 10^{8} m/s}{1.56\times 10^{-18} J}=12.74\times 10^{-8} m=127.4 nm[/tex]
Longest wavelengths of light that can cleave the bonds in elemental nitrogen is 127.4 nm.
Similarly
For oxygen:
Energy to cleave 1 mol O-O bond = 498 kJ/mol = 498000 J/mol
Energy to break 1 O-O bond = [tex]\frac{498000 J/mol}{6.022\times 10^{23} mol^{-1}}=8.26\times 10^{-19} J[/tex]
[tex]E=\frac{hc}{\lambda }[/tex]
[tex]\lambda =\frac{6.626\times 10^{-34}J s\times 3\times 10^{8} m/s}{8.26\times 10^{-19} J}=2.406\times 10^{-7}m=240.6 nm[/tex]
Longest wavelengths of light that can cleave the bonds in elemental oxygen is 240.6 nm.
For fluorine
Energy to cleave 1 mol F-Fbond = 159 kJ/mol = 159000 J/mol
Energy to break 1 F-F bond = [tex]\frac{159000 J/mol}{6.022\times 10^{23} mol^{-1}}=2.64\times 10^{-19} J[/tex]
[tex]E=\frac{hc}{\lambda }[/tex]
[tex]\lambda =\frac{6.626\times 10^{-34}J s\times 3\times 10^{8} m/s}{2.64\times 10^{-19} J}=7.529\times 10^{-7}m=752.9 nm[/tex]
Longest wavelengths of light that can cleave the bonds in elemental fluorine is 752.9 nm.
