## Usuing Rydberg's Equation Conceptually

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Raphael_SanAndres3C
Posts: 33
Joined: Fri Sep 28, 2018 12:24 am

### Usuing Rydberg's Equation Conceptually

For Rydberg's equation, I have seen in previous threads that the negative means that the there is a decrease in energy. So, should I proceed with the problem with the negative from the formula or add the negative in the end. For example, if I am trying to find the energy emitted from when the level of the electron decreases from n=3 to n=1, should I remove the negative from - hR/(n^2) and simply do EnergyFinal(n=1) - EnergyInitial(n=3) then add the negative once I have done calculations or include the negative and have an equation which looks like EnergyProton = (-hR/(1^2)) - (-hR/(3^2)).
I assume the answer would be the same?

Also, as a clarification, when the atom is emitting energy, the answer would be a negative and when the atom is absorbing energy, the answer would be a positive?

Chem_Mod
Posts: 17949
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 406 times

### Re: Usuing Rydberg's Equation Conceptually

1st clarification: We are not using Rydbergs equation.

We use E= $-\frac{hR}{n^{2}}$
The negative is always in this equation which tells us the energy value of each energy level.

Emission of a photon means higher to lower energy level which is a negative deltaE but the photon energy is always positive.