help on 1.15 6th edition

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805098281
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Joined: Fri Sep 28, 2018 12:26 am

help on 1.15 6th edition

Postby 805098281 » Fri Oct 19, 2018 4:20 pm

1.15 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'm so confused, can someone please tell me where to start and where to go from there?

Samantha Kwock 1D
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

Re: help on 1.15 6th edition

Postby Samantha Kwock 1D » Fri Oct 19, 2018 4:40 pm

The ultraviolet spectrum, or the Lyman series, involves electron transitions from a higher energy level to n=1. The wavelength of the emitted light can be converted into frequency. This frequency can then be used in conjunction with Rydberg's equation to determine the value of the second principal quantum number knowing that n1=1.

jguiman4H
Posts: 31
Joined: Fri Sep 28, 2018 12:29 am

Re: help on 1.15 6th edition

Postby jguiman4H » Fri Oct 19, 2018 4:52 pm

They want you to find the initial and final values of "n", so you know you'd need the Rydberg equation.

But, in order to use the Rydberg equation, we'd use the frequency.

So, because we're given 102.6 nm, that equals 102.6 x 109 m. And plugging it in to: c = lambda * nu OR speed of light = wavelength * frequency, we'd get frequency equals 2.922 x 1015 Hz.

Plugging this into the Rydberg equation, we'll eventually get 1/n22 = 0.112. From this, we'll get n22 = 9, and square rooting this, n2 = 3.

Thus, as stated above, we already know that n1 = 1. So, the transition is from n1 = 1 to n2 = 3.


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