## finding n values

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

bgiorgi_3A
Posts: 56
Joined: Wed Sep 30, 2020 9:50 pm

### finding n values

Hey guys! For some reason I'm really struggling with setting up this problem. I thought I knew what I was doing, but I keep getting the answer wrong and so I'm looking for some help!

Q7. An electric current is passed through a tube that contains Hydrogen gas. Light is then passed through the prism and the emission spectrum of Hydrogen gas is recorded. A band was recorded, and it has a frequency of 6.1684E14 hz. What is the value of n initial?

Courtney Situ 2B
Posts: 109
Joined: Wed Sep 30, 2020 10:03 pm

### Re: finding n values

Hi there!

NOTE: After I typed this up, I realized that I technically solved for n2. However, since the light is part of the visible light spectrum, it is part of the Balmer series and n1 should always equal 2. I'm unsure if this is just a typo, but otherwise, I got n2 = 4.

We would start this problem by setting up the Rydberg equation: v = R[(1/n1)^2 - (1/n2)^2]
The next important thing to note is the frequency given (6.1684E14 hz). The band is located within the visible light spectrum, as the visible light spectrum ranges from about 4 x 10^14 Hz ~ 8 x 10^14 Hz. Using this information, we can establish that this band is a part of the Balmer series and n1 = 2. (This is part of the definition of the Balmer series. The bands in the Balmer series all share n1 = 2. For the Lyman series, all the bands share n1 = 1.)

We plug in these values into the Rydberg equation.
6.1684E14 hz = 3.28984487239343E+15 hz * [(1/2)^2 - (1/n2)^2]
When you solve for n2, you should get 3.9999, so I think it's safe to say the answer is 4.

I hope this helps!

Marisa Gaitan 2D
Posts: 107
Joined: Wed Sep 30, 2020 9:47 pm
Been upvoted: 3 times

### Re: finding n values

To solve this you would use the v=R(1/n1^2-1/n2^2) equation where v is frequency and R is rydberg's constant. Before solving, you have to figure out the final energy state. You can do this by converting the frequency 6.1684x10^14Hz to wavelength to get 486.35nm. This is in the visible region and therefore is a part of the Balmer series. Because the Balmer series has a final energy state of n=2, you can then substitute 2 for n1 in the equation, do some algebra and get n2^2=16, so n=4. Hope this helps!

Kayla Law 2D
Posts: 101
Joined: Wed Sep 30, 2020 10:01 pm

### Re: finding n values

Hi! You can also solve this problem using the E = -hR/n^2 equation. First, I converted the frequency into the change in energy by using E = hv. I got that the change in energy is about 4.08718 x 10^-19 J. Then I solved for n final by converting the frequency, 6.1684 x 10^14 Hz, into wavelength. I did this by taking the energy that is just solved for and plugged it into the equation E = hc/λ. I got a wavelength of 486.35 nm, which falls into the visible region, which shows that this is the Balmer series. Therefore, we know that the n final will be n = 2 since for the Balmer series, the final n is always n = 2. Next, since we know the final n, I plugged in 2 into the E = -hR/n^2 equation for n in order to find out the energy of the electron at n = 2. Since we now have the change in energy and the final energy, we can use the equation ΔE = Efinal - Einitial. Make sure to make the ΔE negative because when doing Efinal - Einitial, the change in energy should be negative since the Efinal is more negative than Einitial, so a more negative number - a less negative number will give you a negative number (kinda confusing). Next, plug in your change in energy and the Efinal that you solved for into the equation and you should end up with an Einitial of -1.3627 x 10^-19 J. Then, just plug that into the E = -hR/n^2 equation and solve for n. I got n = 4. All in all, it takes a lot less time to do this using the Rydberg Equation, but doing it this way helps me understand the concepts better (but during a time constraint I would definitely recommend using the Rydberg Equation). I hope this helps! :)