1A.15

$c=\lambda v$

304976622
Posts: 25
Joined: Wed Nov 15, 2017 3:01 am

1A.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 got 2.9x10^15=3.28x10^15{1/1^2-1/n^2} (is this right?)

I understand that we have to solve for n^2 however I'm confused on how to enter this problem into my calculator. Where do I start?

Cynthia Rodas 4H
Posts: 51
Joined: Wed Sep 18, 2019 12:21 am

Re: 1A.15

Yes! Great start so far. From here, you would try to get $\frac{1}{n_{2}^{2}}$ by itself by moving all other values to the other side to get:
$\frac{-2.922 * 10^{15}}{3.29 * 10^{15}} + 1 = \frac{1}{n_{2}^{2}}$

From here, you would solve the left side which is approximately:
$0.11185 = \frac{1}{n_{2}^{2}}$

Then you would get n22 by itself to get:
$n_{2}^{2} = \frac{1}{0.11185}$
which is approximately 9.

Next, you would find the square root to get n2 which is 3.

304976622
Posts: 25
Joined: Wed Nov 15, 2017 3:01 am

Re: 1A.15

Cynthia Rodas 4H wrote:Yes! Great start so far. From here, you would try to get $\frac{1}{n_{2}^{2}}$ by itself by moving all other values to the other side to get:
$\frac{-2.922 * 10^{15}}{3.29 * 10^{15}} + 1 = \frac{1}{n_{2}^{2}}$

From here, you would solve the left side which is approximately:
$0.11185 = \frac{1}{n_{2}^{2}}$

Then you would get n22 by itself to get:
$n_{2}^{2} = \frac{1}{0.11185}$
which is approximately 9.

Next, you would find the square root to get n2 which is 3.

Great thank you!!!

Hailey Kim 4G
Posts: 110
Joined: Sat Jul 20, 2019 12:16 am

Re: 1A.15

How do you know that n1=1?