A light source's maximum wavelength at which o2 can ionize is 99.39nm.
How is calculated to maximum wavelength?
- The energy difference between these two energy levels is equal to the energy needed for this transformation.
- The data can be studied and displayed to find the wavelength of greatest extinction. As a result, the wavelength of light needed to excite the pi electrons to their first excited state is 43.948 nm.
O2 has a 1205 kJ ionization energy.
Energy = hv
= h light's wavelength/velocity
wavelength = (1205000 3 108) /6.626 1034
99.39 nm is the wavelength.
- He is symbolized by the highest ionization energy.
- The outermost shell has a high ionization energy, is stable, and does not frequently become unstable due to electron loss. Ionization can be induced by waves with energies greater than 134 nm.
- 225nm light lacks the energy to ionize gold because it is greater than 134nm.
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