Achieve HW #11

[905767718]

I'm working on a problem where I need to determine whether a spectrum line of wavelength 434.0 nm is Lyman or Balmer series, and I don't know how to figure that out. I already divided the frequency of the light by Ryder's constant giving me 0.178 but I don't know what to do with that number.

[Austin_Schwartz_1L]

For this problem, you must have the information memorized that the Lyman series is in the ultraviolet range while the Balmer series contains the visible light range. The problem gives you a wavelength of 434 nm, which is contained in the visible light region, and therefore falls into the Balmer series. This means that n1 must be n=2. Next, set up the Rydberg Equation to solve for n2. (v=R[(1/n1^2)-(1/n2^2)].

[Erin Chin 1L]

Hi! Well first, to determine whether the spectrum line belongs to the Balmer or Lyman series, you must know that the Lyman series consists of a series of lines in the ultraviolet region while the Balmer series consists of a series of lines in the visible region of the spectrum. Since the violet line from the given problem is in the visible region, it belongs to the Balmer series, and therefore n1=2 (if it was part of the Lyman series, n1=1). To find the n2 energy level, you would use the Rydberg equation, and also make sure to convert the wavelength of the violet line to meters.

[Zara Okeiyi 3C]

How do you find online solutions for the textbook problems?

[905767718]

How do you find online solutions for the textbook problems?

Scroll down on the table of contents on the left hand side of the E-book where it says Back Matter. Once you click on Odd Numbered Exercises, you'll have the answers to all the odd questions in the book.

[Ginny Ghang 1B]

According to the atomic spectrum, the Balmer series is a series of lines in the visible region where n1 = 2, and the Lyman series consists of a series of lines in the UV region where n1 =1. The wavelength of the line that they give you is in the visible region, so you know that it is a part of the Balmer series and that you need to use n1 = 2.

[Sevde Coban 2J]

So when both ns aren’t given, do you always assume n1 is always the first value of n given in the series you use? For example, in the Balmer series we are told that n1 =2, so in problems where the Balmer series is used, is n1 always going to be 2?

[Alex_Lee_1K]

I'm working on a problem where I need to determine whether a spectrum line of wavelength 434.0 nm is Lyman or Balmer series, and I don't know how to figure that out. I already divided the frequency of the light by Ryder's constant giving me 0.178 but I don't know what to do with that number.

For this you would want to substitute frequency with (c*h)/λ and then fill in λ with 434x10^-9 in order to convert the nm into m. After this you would divide by Rydberg's constant.

I'm not sure if you said if it was Lyman or Balmer, but if it is UV light it would be Lyman which is n=1 and if its visible light you would use balmer series which is n=2. I believe I saw this same question somewhere else, but you would substitute it for n and then use simple algebra to solve the other n.

Hope this helped.