## Series and Wavelengths part 2

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

Posts: 39
Joined: Fri Sep 28, 2018 12:16 am

### Series and Wavelengths part 2

In another question 1.57, it gives lines in the Balmer series of the hydrogen spectrum (656.3, 486.1, 434.0, and 410.2 nm) and asks for the next wavelength. In the solutions manual is shows that we must solve for energy using n(1) = 2 and n(2) = 7. How come we use n(2) = 7 when it states that its in the Balmer series? Does this somehow gives us n = 6 when solved for the next wavelength?

Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

### Re: Series and Wavelengths part 2

The Balmer series has lines with n(1) = 2 and n(2) > 2. It consists of lines that go from n = 2 as well as value anything above that as the second n value.
So we know:
656.3 is n = 3 --> n = 2
486.1 is n = 4 --> n = 2
434.0 is n = 5 --> n = 2
410.2 is n = 6 --> n = 2
So now the next one would be :
? nm is n = 7 --> n = 2
which can be found using Rydberg's constant and frequency formula. If you have the 6th edition textbook, it's on Section 1.3, page 7.