# Question #07aa9

##### 1 Answer
Jun 6, 2017

They emit a photon on energy equal to the energy difference between the states (from making transition from higher to lower energy state).

Thus if ${E}_{2}$ represent energy of higher level and ${E}_{1}$ represent energy of lower level, then for making a transition from the higher to the lower stationary state, the atom emits a single photon of energy,

$\nu = \frac{{E}_{2} - {E}_{1}}{h}$ where, $h$ is the Planck's constant.

And they do not use that energy as they are excited to the higher level. They retain the energy. There are no dissipative means in which the energy can be dissipated or used up.

As long as the electron stays in the higher energy state, they retain the same energy. Just when they transition to a lower state, the emit the extra energy.

If they would use up energy during their motion in the atom, the stability of matter couldn't not be explained.

That's the most important reason for the failure of the Rutherford's model, where the electrons orbital energy was supposed to be used up in emitting continuous radiation in accordance with Electromagnetic theory.