Photolysis of nitrosobenzene at 248 nm leads to the formation of phenyl radicals.
So to know the phenyl radical concentration, information about the quantum yield is of great importance.
Thank you Sir.
But I have a small doubt that whether the quantum yield changes with the photolysis beam energy. If so in the paper sent by you, the energy is 8 mJ/pulse for the determination of quantum yield where as in my experiments I am using energies in order of 30-40 mJ/pulse.
Can you clarify my doubt?
Parth
Unless the radical formation is a two photon process, the quantum yield for your radical formation should be independent of the photon flux (concentration of photons). Things like the lifetime of your radicals may vary due to increased recombination, but if the correct equation is the simple formula: QYradicalformation = kradical/(kr+knr+kradical) this is independent of [C5H5NO*] there should be no laser power dependence. But if you are concerned you may just half the power and make sure your radical product is consumed by an order of 1/2, 30-40mJ is a lot of power so this may be an important experiment to run if you haven't already.
One issue is that you may have is that your excitation wavelength is different (and higher energy) than the above paper. It may be an important to determine the excited state manifold you are exciting into vs the paper above and consider if the radiation you are using may be ionizing.
Tyler
Thank you for your answer.
No the 248 nm radiation used by me is not going to ionize the nitrosobenzene it is just going to form the radicals. (literature supports this argument and nitrosobenzene is a well known precursor for the formation of phenyl radicals).
I will exercise as to what you have told.
Thank you Sir.
Parth
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