## Exercise 1.15 (Ch 1 : The Quantum World)

Johana Jeon 1A
Posts: 30
Joined: Sun Feb 12, 2017 3:00 am

### Exercise 1.15 (Ch 1 : The Quantum World)

In 1.15: In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6nm. 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.

Question: after getting 0.112, how do we know that n2^2 = 9; n2 = 3?

Chels Zh 1D
Posts: 41
Joined: Fri Jun 23, 2017 11:39 am

### Re: Exercise 1.15 (Ch 1 : The Quantum World)

Hi there, just got a quick thought of what you may concern:
I will follow up the steps next, since you've already got to 0.1112, which is the frequency we need for n1, and by using Rydberg equation we can solve for n2:
v=R(1/n1^2-1/n2^2), from that equation, we have known that constant R = 3.29E15, then direct substitute n1^2, which equals 1, and then calculate the value of n2. I will post a picture of the detailed step of maths below. Please comment below if I got any steps confusing or incorrect and I hope it helps you. ;)
[attachment=0]IMG_0395.JPG[/attachment]
Attachments Johana Jeon 1A
Posts: 30
Joined: Sun Feb 12, 2017 3:00 am

### Re: Exercise 1.15 (Ch 1 : The Quantum World)

Thank you!!

604744616
Posts: 34
Joined: Sun Jun 25, 2017 3:00 am

### Re: Exercise 1.15 (Ch 1 : The Quantum World)

I am still confused to how we go from 0.112 to n^2=3. I am not understanding on how we got the second value of n just from solving of the rydberg equation?

Johana Jeon 1A
Posts: 30
Joined: Sun Feb 12, 2017 3:00 am

### Re: Exercise 1.15 (Ch 1 : The Quantum World)

When you do 1/0.112 it's 8.928571429 which is roughly about 9
Square root of that would be 3