Textbook problem 1.A.15

[Lara Almeida 3D]

PROBLEM:
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.

MY ANSWER:

because the spectral line is in the Uv spectrum it means that n2 = 1

After figuring out the frequency I just plugged the numbers in the Rydenberg equation.

My answer seems to be wrong, could someone please help me identify the reason? Tks!

[Isabel Carden 1E]

Hi!

I was also confused on this question. I followed your exact same steps and came out with the same answer, but the book says that the shells are n=1 and n=3. I tried doing the problem in reverse by plugging in n=1 and n=3 as follows:

v=3.29x10^15(1/1^2-1/3^2)

This got me the same value for the initial frequency that we calculated at the beginning of the problem, which confirms that 1 and 3 are the correct answers, but I'm still confused about how that math isn't checking out in the forward direction. Additionally, when you plug the equation into a solve function on a graphing calculator, you don't get 3 for n, but the instead the 0.727 that we know isn't correct.

[RossLechner3E]

Make sure to be careful with which value is n1 and which is n2. Although you correctly determined the electron is excited to n=1 and although this may be seen as the "final" compared to the initial state, this value is n1, not n2. If you instead try v=R-(R/n^2) then -v+R=(R/n^2) that should yield the correct value of n2=3