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### The Negative Sign in En = -hR/n^2

Posted: **Wed Apr 18, 2018 8:11 pm**

by **Steven Luong 1E**

Hello, the professor mentioned the negative sign in the aforementioned formula is due to the falling of the electron from a higher energy to a lower energy level in which energy is released as photons. Is this correct, or am I missing something as to why there exists a negative sign?

### Re: The Negative Sign in En = -hR/n^2

Posted: **Wed Apr 18, 2018 8:37 pm**

by **Jack Martinyan 1L**

The negative sign implies that the bound electron has lower energy than the free electron. The eletron's energy has decreased, and the energy is released as electromagnetic radiation

### Re: The Negative Sign in En = -hR/n^2

Posted: **Wed Apr 18, 2018 8:49 pm**

by **Salena Chowdri 1I**

That is correct. The energized electron falls from a higher energy level to a lower energy level (releasing photons). The release of energy causes the change in energy (Delta E) to be a negative value.

### Re: The Negative Sign in En = -hR/n^2

Posted: **Sun Apr 22, 2018 12:06 pm**

by **Emma Leshan 1B**

So as a follow up, when using the Rydberg equation, the negative is there for the same reason? Is it because the Rydberg equation is derived from this equation? When calculating frequency from the Rydberg equation, we could just ignore the negative, correct?

### Re: The Negative Sign in En = -hR/n^2

Posted: **Sun Apr 22, 2018 2:19 pm**

by **Betty Wolkeba section 1L**

Why is it when we solve for the E in the Rydberg equation, we have to switch the sign from a negative to a positive when solving for another variable such as wavelength?

### Re: The Negative Sign in En = -hR/n^2

Posted: **Sun Apr 22, 2018 3:01 pm**

by **Chem_Mod**

If you start from the principles of conservation of energy, (E_{total,initial} = E_{total,final}) the rise of negative signs in the formulae will be clear.

For absorption, E_{photon} + E_{initial} = E_{final}. E_{photon} comes in initially, since light comes and hits the atom for absorption. For emission, E_{initial} = E_{final} + E_{photon} . Plug in the appropriate expressions for each terms and you will not have to worry about the negative signs.