The ionization energy for the hydrogen atom is 13.6 eV.What is the energy of a photon that is emitted as a hydrogen atom makes a transition between the n = 4 and n = 2 states?
A)0.85 eV
B)2.55 eV
C)3.40 eV
D)6.80 eV

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The Answer of The ionization energy for the hydrogen atom is 13.6 eV.What is the energy of a photon that is emitted as a hydrogen atom makes a transition between the n = 4 and n = 2 states?
A)0.85 eV
B)2.55 eV
C)3.40 eV
D)6.80 eV

The Ionization Energy for the Hydrogen Atom Is 13