Question

The figure shown above shows the energy levels for a hydrogen atom. What is the energy of the photon emitted when the electron in a hydrogen atom drops from energy level E5 to energy level E2?

Answer 1

The energy of the photon emitted when the electron in a hydrogen atom drops from energy level E5 to energy level E2 is 2.85eV

Now, the energy of the ground state is -13.6 eV.
the energy of the first state (n = 1) is -3.39 eV.
the energy of the fifth state is -0.54 eV.

Now, When an electron drops from higher to lower state, then energy difference is same as the energy of the photon emitted, is
 = E5 - E2
= -0.54 eV -(-3.39 eV)
= 2.85 eV 

Answer 2

The energy involved in the drop of the electron is obtained from  ''ΔE= E5 - E2.''

What is Bohr model?

According to the Bohr model, energy is emitted when an electron drops from a higher to a lower energy level. In this case, the electron has dropped from E5 to E2.

The specific figure is not shown here but we can calculate the energy involved from ''ΔE= E5 - E2.''

Learn more about Bohr: https://brainly.com/question/15167411