## Negatives and Planck's Constant in finding the energy of a transitioning hydrogen atom?

$c=\lambda v$

KC Navarro_1H
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### Negatives and Planck's Constant in finding the energy of a transitioning hydrogen atom?

I know that E_n = -hR/n^2 , and h = Planck's Constant but in the module video for Atomic Spectra, the h is taken away? Also, when you try to find the frequency or wavelength of radiation generated from the transition, are they supposed to remain negative or does the negative just go away?

The example problem from Monday involved a negative change in energy from level 4 to 2, and when we used E = hv to find the frequency, where did the negative go?

JooHyun Koh 1H
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### Re: Negatives and Planck's Constant in finding the energy of a transitioning hydrogen atom?

Energy can never be negative; however, delta E, can be. Delta E is what we get when we try to figure out what the change in energy is when an electron jumps energy levels. Energy of the photon emitted during that process is positive, because according to the conversation of energy, no energy can be created or destroyed. So, to follow this rule, the energy of the photon emitted is possible, to compensate for the negative change in energy that occurs in jumping down energy levels.

Chem_Mod
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### Re: Negatives and Planck's Constant in finding the energy of a transitioning hydrogen atom?

Sometimes, like in the Rydberg equation, $\nu = R_{H}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )$, we have divided both sides of the equation by h.

The negative sign tells us whether it's absorption or emission. If the change in energy is positive, a photon was absorbed, and n increased. If the change is negative, a photon was emitted, and n decreased. But in either case, we need the absolute value of the change in energy when we use $E=h\nu$ to calculate the frequency. This is why it sometimes seems like we've dropped the negative sign.

If you write out the equation for conservation of energy for the transition, the negative sign will automatically be resolved:
Absorption:
(energy of photon) + (initial energy of hydrogen atom) = (final energy of hydrogen atom)
$h \nu=E_{final}-E_{initial}$
Emission: (initial energy of hydrogen atom) = (final energy of hydrogen atom) + (energy of photon)
$h \nu=E_{initial}-E_{final}$
In both cases the right side of the equation is positive.