## Problem A.15 values of n

$c=\lambda v$

kevinchang_4I
Posts: 55
Joined: Tue Nov 13, 2018 12:17 am

### Problem A.15 values of n

In this problem it gives us the wavelength of 102.6nm and tells us the line is in the UV spectrum. This means that n1 is going to = 1, and we must find n2. v=R((1/n1^2)-(1/n2^2)), and wavelength 102.6nm = c / v. I just don't know the algebra to find the value of n2. Thanks for the help

DMuth_1J
Posts: 63
Joined: Thu Jul 11, 2019 12:15 am

### Re: Problem A.15 values of n

Algebraically you could isolate n2^2 in either formats

(1/n2^2)= (1/n1^2)-(V/R)

or

(n2^2)= (n1^2)-(R/V)

BNgo_2L
Posts: 95
Joined: Wed Sep 11, 2019 12:17 am

### Re: Problem A.15 values of n

After converting the wavelength of the spectral line to the energy it emits, you have to use E=E(final)-E(initial) to figure out the n values.You're already given the n value for the final energy as UV follows the Lyman series stating from any n value, the electron must return to ground state at n=1. Next, you find the Energy for the final and then plug all of the given information into the change in E equation in order to isolate n for the initial energy. Isolating n of the initial energy will give you the last n value.