## HW 1.15

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Nicole Gamboa 2M
Posts: 21
Joined: Wed Sep 21, 2016 2:56 pm

### HW 1.15

On problem 15 of the homework, it is given that a line is observed at 102.6 nm and asks for you to determine the values of n for the initial and final energy levels. How do you determine n if they are both variables?

Manpreet Singh 1N
Posts: 41
Joined: Wed Sep 21, 2016 2:59 pm
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### Re: HW 1.15

Hello,

The question tells us that a line at 102.6nm was observed in the ultraviolet spectrum. This brings in the Balmer and Lyman series. We know that Balmer is visible light with n1=2, while Lyman is UV with n1=1. I think it is something we have to remember.

we can solve this problem by using v=R(1/n1^2-1/n2^2)

we would solve for frequency by using v=c/$\lambda$=3.0x10^8 m/s /(102.6X10^-9m)=2.922x10^15

2.922x10^15=3.29x10^15 1/s (1/1-1/n2^2)

from there you just need to solve for n2

hope that helps :)

danae_blodgett_1H
Posts: 18
Joined: Sat Jul 23, 2016 3:00 am

### Re: HW 1.15

How would we solve this problem using the formula we used in class En=-(hR/n^2)? He never went over this formula in class so won't you get the same answer using the other formula as well?

Chris_Rudewicz_3H
Posts: 32
Joined: Fri Jul 22, 2016 3:00 am
Been upvoted: 1 time

### Re: HW 1.15

danae_blodgett_3F wrote:How would we solve this problem using the formula we used in class En=-(hR/n^2)? He never went over this formula in class so won't you get the same answer using the other formula as well?

The formula in the reply above is the same as the formula we used in class (En = -(hR/n^2)). But the question is asking for the change in energy from one level (n) to the next. Here, because it is Balmer series, the electron goes to energy level n=2 so you are left with only one variable since h and R are constants.

Also, E=hv, so in the reply above he/she simply simplified the equation by dividing everything by h. Thus giving you v = -R/n^2 - (-R/n^2). You can factor out a -R and solve for your unknown n. Hope this helps.