Hydrogen

find the energy?

find the eneregy in eV of a photon that is emitted when an electron in a Hydrogen atom undergoes a transition from the n=7 level and produces a line in the Balmer series.

energy associated with (n) orbit of H2 atom is
En = – 13.6 eV/n^2 where n = 1 2 3 ….
higher is the orbit, more is the energy >> n =infinity entails zero (max) energy
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energy of photon when electron jumps from n1 to n2 orbit =
E = hv = -13.6[1/n1^2 - 1/n2^2] eV
transition from comes n1 =7
E(7) = – 13.6/49 = – 0.2775 eV (higher)
for Balmer, electron is absorbed in n2 =2
E(2) = – 13.6/4 = – 3.4 eV (lower)
photon E = – 0.2775 – (- 3.4) = 3.1225 eV
E = 3.1225*1.6*10^-19 = 4.996*10^-19 Joule