Asked by Kjd

The most prominent line in the atomic spectrum of mercury occurs at a wavelength of 253.652 nm. What is the energy of 0.625 moles of photons associated with the most prominent line? Express answer in kJ.

ANS:
295 kJ

Please explain.. Thank you so much!
Answered by DrBob222
E = hc/wavelength
Solve for E. Wavelength must be in m and c in m/s. E is in joules/photon; therefore, multiply by 6.02E23 to convet to a mole of photons and multiply that by 0.625 for 0.625 mols photons. The answer is in joules.
Answered by Kjd
Got it, thank youu!
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