## hw problem 1.A.15

$c=\lambda v$

Bita Ghanei 1F
Posts: 60
Joined: Thu Feb 28, 2019 12:15 am

### hw problem 1.A.15

In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line.

Could someone please walk me through the math here? Thanks!

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

### Re: hw problem 1.A.15

To solve this problem, we'll be using $\lambda*v = c$ and then Rydberg's equation $v = \left (1/n_{2}^{2} - 1/n_{1}^{2}\right )*R$.
The info we have here is the wavelength and the initial level of the electron (n=1 since we know that the H atom emitted energy to reach a new energy level where $\lambda = 102.6nm$).

The first step is finding the frequency:
$\lambda * v = c$
$1.026*10^{-7} nm *v = 3*10^{8} m$
$v =2.92*10^{15} Hz$

Then, we can use the frequency to discover what n(final) or $n_{2}$ is:
$v = \left (1/n_{2}^{2} - 1/n_{1}^{2}\right )*R$
$2.92*10^{15} Hz = \left (1/n_{2}^{2} - 1/n_{1}^{2}\right )*3.29*10^{15}$
$0.888 = 1/1^{2} - 1/n_2^{2}$
$n_2^{2} = 8.93$
$n_2 = 2.99 = 3$

Alexa Mugol 3I
Posts: 54
Joined: Sat Aug 17, 2019 12:17 am

### Re: hw problem 1.A.15

The previous person explained it pretty well! I'd like to add that we know that it starts at energy level 1 (n1=1) because it's in the ultraviolet spectrum, aka the Lyman series, so it's starts at n=1.

JChen_2I
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am

### Re: hw problem 1.A.15

Just to clarify, we sub in 1 for n2 because n2 is the final energy level right? And n1 is the initial energy level? So it should be final n subtracted by the initial n