Question

In a simplified version of the solar system model (developed by Niels Bohr, a Danish chemist), an electron resides in an excited H atom orbit with an energy of -38.42 J (initial state). If the electron drops to a second orbit (final state) with the energy -63.33 J, a photon will be emitted. What will be the energy of the photon (J)?

can someone show how to solve this problem and explain.

Answer 1

-24.91 J

Step-by-step explanation:

Energy of emitted Photon = Energy of final state - Energy of initial state

= -63.33 -(-38.42)

= -24.91 J

The negative sign indicates that a photon was emitted.