Answer :
The maximum wavelength of light capable of causing the ionization of o2 is 99.39 nm.
The energy required for this transition corresponds to the energy difference between these two energy levels. Therefore, the wavelength of light required to bring the pi electrons into their first excited state is 43.948 nm, and the data can be examined and plotted to determine the wavelength of maximum extinction.
calculation:-
ionization energy of O2 = 1205 kJ
energy = hv
= h × wavelength/speed of light
wavelength = 1205000× 3 ×10^8) /6.626 ×10^34
wavelength = 99.39 nm.
The highest ionization energy represents Him. Because the outermost shell is completely filled it is stable does not tend to become unstable due to electron loss, and has high ionization energy. Wavelengths shorter than 134 nm have enough energy to promote ionization. Since 225nm is larger than 134nm 225nm light does not have enough energy to ionize gold.
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