## Ways of Doing Problem 1.15?

Alex Nguyen 3I
Posts: 100
Joined: Fri Sep 25, 2015 3:00 am

### Ways of Doing Problem 1.15?

"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." I know you can do this by solving for frequency (nu) and solving the Rydberg Formula for n sub 2, but can you use the equation for energy change? Change in E= E final- E initial, where E= -hR/n squared.

Alan Chien 1J
Posts: 72
Joined: Fri Sep 25, 2015 3:00 am

### Re: Ways of Doing Problem 1.15?

Although not the most efficient way, I personally calculated the energy levels of n=1 through 5 which surprisingly does not take that much time and compared the differences until I found one that matched the energy emitted given the wavelength was 102.6 nm.

Alex Nguyen 3I
Posts: 100
Joined: Fri Sep 25, 2015 3:00 am

### Re: Ways of Doing Problem 1.15?

Well actually, you could just solve the Fundamental Energy Difference Equation for n2. Change in E= $\frac{-hR}{n^{2}}+\frac{hR}{n^{2}}$. Just remember that it's Final Level- Initial Level. We know that the final level is 1 because it's in the Lyman Series for UV EM radiation. You're given the wavelength, so you can find the energy in Joules and set up the equation to equal this. The only trick part to this is that you must use the negative value of the energy, and that's simply because the fundamental equation for E given n places 0 as the reference point (where the electron is ionized). Energies below ionization are negative.