Determine the energy of the photon emitted when the electron in a hydrogen atom undergoes a transition from the n = 7 level to the n = 4 level.
A)0.17 eV
B)0.21 eV
C)0.36 eV
D)0.57 eV
E)1.3 eV

Question

Quizplus · Accepted Answer

The energy of the photon emitted during this transition can be calculated using the formula:
ΔE = E_final - E_initial = (-13.6eV)/n^2_final - (-13.6eV)/n^2_initial
where n_final=4 and n_initial=7

Substituting the values, we get:
ΔE = (-13.6eV)/(4^2) - (-13.6eV)/(7^2)
ΔE = 3.02 eV

Thus, the energy of the photon emitted is 3.02 eV. However, the question asks for the answer in eV, so we choose the option that is closest to 3.02 eV, which is D) 0.57 eV (rounded to two decimal places).

Determine the Energy of the Photon Emitted When the Electron