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

Steven Luong 1E
Posts: 33
Joined: Fri Apr 06, 2018 11:03 am

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?

Jack Martinyan 1L
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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

Salena Chowdri 1I
Posts: 66
Joined: Fri Apr 06, 2018 11:02 am

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.

Emma Leshan 1B
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Joined: Fri Apr 06, 2018 11:03 am

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?

Betty Wolkeba section 1L
Posts: 32
Joined: Fri Apr 06, 2018 11:05 am

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?

Chem_Mod
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