Balmer Lines

[UrbanXrisis]

I'm learning about quantum physics and did an experiement in my class. It's a spectroscopy activity where you place a hydrogen lamp in front of a defraction grating. We must verify that only the Balmer lines are visible but why do I only observe the Balmer series lines and not others?

[Dr Transport]

Because the Balmer series is in the visible wavelength region.....

[ZapperZ]

I'm learning about quantum physics and did an experiement in my class. It's a spectroscopy activity where you place a hydrogen lamp in front of a defraction grating. We must verify that only the Balmer lines are visible but why do I only observe the Balmer series lines and not others?

Only the Bamer lines are in the visible range for your eyes. The Lyman series, for example, are in the ultraviolet range, while the Paschen series are in the infrared. You can verify this by looking at the wavelength typical for each series. All these lines are probably there (depending on the power supply attached to your discharge tubes), but you just can't see them.

Zz.

[UrbanXrisis]

Sorry to bore you. Like I said, I'm totally new at this. The experiment asks me to verify that only the Balmer lines are visible with the equation:

1/lambda=R(1/n^2)-(1/n^2)

[nickdanger]

Sorry to bore you. Like I said, I'm totally new at this. The experiment asks me to verify that only the Balmer lines are visible with the equation:

1/lambda=R(1/n^2)-(1/n^2)

So then take the n values (principle quantum number) for just the balmer series. I think they are n=[3,4,5,6...]. But you simply go back and look up in the textbook (or web) that they are visible wavelengths that result from putting these numbers in the above Balmer formula....right? All other series (like Paschen) will calculate to be outside your vision. Now the tough part is arranging your equation properly. It says above there are two n's, but its easier to think of them separately as m and n where m<n. We will hold m=2 as constant for the whole series (that's what defines the Balmer series) and increment the n to get all the lines. So the first will be m=2 and n=3.

1. R[1/m^2 - 1/n^2] = R[1/4 - 1/9] = 1.09723 10E-3[0.13888] = 1/(6562.08 Angstroms)

and that's correct. You are using the Rydberg constant above. Then,

2. R[1/4 - 1/16] = 1/(4860.7 A)

Is that right? I'll bet all the lines you saw in your lab correspond to everything you calculate with this method. Now keep going. I don't know how many you calculate for all the visible lines, but as I said you can look that up