Chapter 1.15 6th Edition


Moderators: Chem_Mod, Chem_Admin

chrisdinkel_4E
Posts: 33
Joined: Fri Sep 28, 2018 12:17 am

Chapter 1.15 6th Edition

Postby chrisdinkel_4E » Thu Oct 25, 2018 10:12 pm

this question covers the rydberg equation. in the solution, it assumes that n(final) is 1 and uses that assumption to find that n(initial) is 3.
Can someone explain if this assumption is always valid? Thanks

Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

Re: Chapter 1.15 6th Edition

Postby Andrea Zheng 1H » Thu Oct 25, 2018 10:15 pm

this question says the wavelength is equal to 102.6 nm. This wavelength corresponds to the Lyman series, a set of lines in the UV region of the spectrum with n1 = 1. This is why the answer uses the assumption that n1 = 1. If the wavelength corresponded to the Balmer series, then n1 = 2.

Cade Okohira 4K
Posts: 47
Joined: Fri Sep 28, 2018 12:15 am

Re: Chapter 1.15 6th Edition

Postby Cade Okohira 4K » Thu Oct 25, 2018 10:51 pm

We know that the since the wavelength is 102.6 nm, the n(1) must be equal to 1 because it is part of the Lyman series. However, to find the initial value of n, we must use Rydberg's equation.

First, we can use v=c/wavelength to find the frequency of the electron: v=(3x10^8 m/s)/(102.6x10^-9 m) = 2.924x10^15 Hz.
Now that we have the frequency, we can use Rydberg's equation and solve for the inital n:
v=R((1/1^2)-(1/n^2)) --> 2.924x10^15 Hz= R((1/1^2)-(1/n^2))

After doing some algebra, you can find that n^2=8.98, which is very close to 9. Thus, the square root of n^2 is 3, so this is the initial energy level.


Return to “Properties of Light”

Who is online

Users browsing this forum: No registered users and 1 guest