The ionization energy of the hydrogen atom is 13.6 eV.What is the energy of a photon emitted corresponding to a transition from the n = 5 to n = 2 state?
A)2.9 eV
B)3.5 eV
C)4.0 eV
D)7.9 eV

Question

Quizplus · Accepted Answer

The energy of a photon emitted during a transition from energy level n1 to n2 is given by: E = (hcR)/((1/n1^2) - (1/n2^2)), where h is Planck's constant, c is the speed of light, and R is the Rydberg constant. Substituting the given values, we get:
E = (6.626 x 10^-34 J s x 3 x 10^8 m/s x 1.097 x 10^7 m^-1)/((1/5^2) - (1/2^2))
E = 2.86 x 10^-19 J
Converting to eV: E = (2.86 x 10^-19 J)/(1.602 x 10^-19 J/eV) = 1.79 eV
Thus, the energy of the emitted photon is 1.79 eV, which is closest to choice A (2.9 eV).

The Ionization Energy of the Hydrogen Atom Is 13