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 = 3 states?
A) 3.40 eV
B) 0.66 eV
C) 6.80 eV
D) 2.55 eV

Question

Quizplus · Accepted Answer

The energy difference between levels n=4 and n=3 in a hydrogen atom is calculated using the formula: E = 13.6 eV * (1/n1^2 - 1/n2^2). For n1=3 and n2=4, E = 13.6 * (1/3^2 - 1/4^2) = 0.66 eV.

The Ionization Energy for the Hydrogen Atom Is 13