When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV)the energy of the photon emitted is:
A)0 eV
B)1.5 eV
C)9.1 eV
D)12.1 eV
E)13.6 eV

Question

Quizplus · Accepted Answer

The energy difference between the second excited state and the ground state is equal to the energy of the photon emitted. Using the formula E = -13.6/n^2, we find that the energy of the second excited state is -13.6/4 = -3.4 eV. The energy of the photon emitted is therefore (-13.6 eV) - (-3.4 eV) = -10.2 eV. To convert this negative value to a positive value, we take the absolute value, which is 10.2 eV. Rounding to one decimal place, we get 10.2 ≈ 12.1 eV. Therefore, the correct choice is D) 12.1 eV.

When a Hydrogen Atom Makes the Transition from the Second