## Wavelength question 1.57 (6th edition)

$\lambda=\frac{h}{p}$

taiyeojeikere-3C
Posts: 30
Joined: Fri Sep 28, 2018 12:26 am

### Wavelength question 1.57 (6th edition)

Is there a calculation to solve the next number in a wavelength in a Balmer series or???
Question: Lines in the Balmer series of the hydrogen spectrum are observed at 656.3, 486.1, 434.0, and 410.2 nm. What is the wavelength of the next line in the series?

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

### Re: Wavelength question 1.57 (6th edition)

So you know that the equation for the hydrogen spectrum is En = -hR/n^2. where R = 3.29E15 Hz. In this case you know that the Balmer series is where there is a transition from any energy level greater than 2 to n = 2. So 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. Next, you expect the next line to be n = 7 --> n = 2.
So En = n(final) - n(initial) --> En = (-hR/2^2) - (-hR/7^2) = -5.0E-19J
The negative changes to positive in the next calculation to find wavelength because that amount just shows how much energy has been emitted/released.
E = hc/λ --> λ = hc/E
Plug in the energy you got and get your wavelength.