Steve C. answered • 06/10/15
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Steve C. Math & Chemistry Tutoring
This problem can be solved using Planck's equation : E = hν = hc/λ:
Ephoton = (6.626 x 10-34 J-sec) (2.9979 x 108 m/sec) / (121.5 x 10-9 m) = 1.6349 x 10-18 J
Emol = (1.6349 x 10-18 J/photon)(6.022 x 1023 photons/mol) = 984539 J = 984.5 kJ / mol
#1 is the answer
Steve C.
From the answer choices it is apparent that the problem asks for the energy involved in a total of a mole of electrons dropping from the second energy level of the hydrogen atom to the first energy level. If the answers were listed as just joules (and the values were very small) then one would give the energy involved for just one electron.
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06/11/15
Aanchal C.
I have understood it now ..thank you sir.
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06/11/15
Aanchal C.
i get the first part E photon. , i solved my question till here, but after this step how does Emol be equivalent to the energy difference between first and second shells of hydrogen atom , i am little confused.
06/11/15