## test

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

Jimmy lira-1G
Posts: 61
Joined: Fri Apr 06, 2018 11:03 am

### test

for the test we took there was a problem that gave us the type of light and said to find resting and excited states of n and calculate n , how would this be solved without knowing the value of n??

-Jimmy Lira -1g

Mariah Guerrero 1J
Posts: 32
Joined: Fri Apr 06, 2018 11:03 am

### Re: test

You can use Rydberg's Equation to solve for the energy levels (n). E= -hR/n^2
If the energy is given in the problem, you would simply have to rearrange the equation in terms of n as it is the only unknown in this case.

Cindy Nguyen 1L
Posts: 48
Joined: Fri Apr 06, 2018 11:04 am

### Re: test

In my version of the test and if I'm thinking of the same problem, it did say the electrons were excited from a ground state to an excited state. It did not say which energy level it was excited to though. However, you do not need to know the energy level to solve this problem. It gave you a wavelength, and you had to find the energy. All you needed to do is find the frequency from the given wavelength, so you can plug the frequency into E=hv and find the energy.

For an electron to be excited from ground state to another energy level would require a specific amount of energy. Because we are told it was excited, then whatever wavelength the test gave would mean the photon gave enough energy (which was what we were supposed to find).

However, you could find out what energy level it was excited to with all this information, but that would require outside information.

Andrew Evans - 1G
Posts: 47
Joined: Fri Apr 06, 2018 11:02 am
Been upvoted: 1 time

### Re: test

Hi Jimmy,
I don’t know what the problem was, or whether you had to find the energy levels, but I’m just gonna answer your question anyway.

Essentially if you’re given the wavelength, you can find the energy (E=h*c/λ). However to find the energy levels it changes to and from, calculated through ΔE. But you need to assume at least one of the values of n in order to find the other. You tell yourself “WHAT” “how do I do this”. but But BUT. There’s the thing. If you’re given the wavelength and you're given that it is indeed a Hydrogen atom, you can assume something after all! The Lyman series of spectral lines all drop to the same energy level (n=0) and same forthe baller series (n=1). Since the Lyman series covers UV light, if the equation mentions the emitted light is UV light, you can assume it is the Lyman series and that nfinal=0. Similarly if it says the emitted light is visible light, assume it’s the baller series so that nfinal=1.
Then you use that to calculate ninitial.

-Andrew Evans
Section 1G