$H_{\psi }=E_{\psi }$

1-D: $E_{TOTAL}\psi (x)=E_{k}\psi (x)+V(x)\psi(x)=-\frac{h^{2}}{8\pi ^{2}m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$

sbeall_1C
Posts: 109
Joined: Sat Sep 07, 2019 12:17 am
Been upvoted: 1 time

In class and the workshops, we looked at a diagram where the Y-axis is Energy, and the values are all negative building up to E=0 when the energy level goes toward infinity. Can someone explain why the energy values are negative? Is it due to the equation?

Mansi_1D
Posts: 50
Joined: Fri Aug 02, 2019 12:15 am

The values as electrons are going down to E=0 is negative because energy is being released. When energy is released, it is shown as a negative value. As electrons go from a higher energy level to a lower one, a photon (energy) is released.

Megan_1F
Posts: 49
Joined: Thu Jul 25, 2019 12:16 am

Is Rydberg's constant negative? Or is the equation itself just negative?

Quynh Vo
Posts: 45
Joined: Sun Sep 29, 2019 12:15 am

Rydberg's constant is not a negative number.
Megan_1F wrote:Is Rydberg's constant negative? Or is the equation itself just negative?

katrinawong3d
Posts: 40
Joined: Sat Aug 17, 2019 12:17 am

To add on, when electrons move down an energy level, it is an exothermic reaction, which indicates a negative sign.

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