## Dino nuggets 11b

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

Indy Bui 1l
Posts: 99
Joined: Sat Sep 07, 2019 12:19 am

### Dino nuggets 11b

Can someone please explain Dino Nuggets 11.b)

Here is the question for reference "If a hydrogen electron goes from n=6 to n=4, will ∆E be negative or positive? Will the energy of the photon emitted be negative or positive?"
I don't quite understand how to determine if the energy of the photon will be positive or negative.

Justin Seok 2A
Posts: 104
Joined: Sat Aug 24, 2019 12:15 am

### Re: Dino nuggets 11b

Essentially, you can tell whether an electron/photon will be emitted or absorbed based on whether n goes up or down. E for an electron is measured negatively, meaning that when n approaches infinity, E approaches 0. When n=1, the lowest quanta it can be, E becomes -hR/n^2, the most negative value E can be. So when n=6 goes to n=4, the value of E decreases, as E goes from a less negative number to a more negative number as it gets closer to the nucleus. Thus energy is released and a photon is emitted. The energy of the photon will always be positive, as a photon simply can't have a negative value for energy. In the case where n goes from 4 to 6, for example, change in E would become positive, as E goes from a more negative to a less negative value, meaning that energy has been absorbed, most likely from an electron. The energy of a photon, when emitted would always be positive, as the energy lost by the hydrogen electron is used in the form of positive energy of the photon emitted.

Jessica Li 4F
Posts: 115
Joined: Fri Aug 09, 2019 12:16 am

### Re: Dino nuggets 11b

Emission means releasing energy, which means that delta E itself will be negative, as energy is lost.
The energy of a photon, however, can never be negative, so it is always the positive absolute value of delta E.

Indy Bui 1l
Posts: 99
Joined: Sat Sep 07, 2019 12:19 am

### Re: Dino nuggets 11b

Awesome thank you for the responses!