Exercise 1A.15 [ENDORSED]

[Tala Ayoub 1H]

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.

How do you solve this question, I calculated the frequency using the wavelength and then divided it by rydberg's constant and I now have 1/(n2)^2 - 1/(n1)^2 = value which was calculated by freq/rydberg's constant but I don't know where to go from there. Is this correct?

[Julie Tran 3C]

Hello,

You are correct! The bit of information you're missing is that a 102.6 nm emission is UV radiation, suggesting that this emission is part of the Lyman series, meaning that the final energy level the electron ends up in is n = 1. Plug that in for n2 and solve for n1!

For some additional information, if the emission is in the visible light spectrum, it's part of the Balmer series, meaning that the final energy level of the electron is n = 2. These two series should cover most of the calculations you would do for this.

Hope this helps!

[Alekhya_Pantula_2E]

Hello!
So looking at the wavelength of the spectrum line, seeing as it is at 102.6 nanometers and in the ultraviolet spectrum, we can conclude this is part of the Lyman series where the final state of the electron is in the n=1 energy level. From there you can plug in 1 into the Rydberg equation and solve for what the initial n would be.
Hope this helps!

[Chem_Mod]

As posted earlier, an essential part to this question is knowing which hydrogen emission series the question is referring to. Knowing that the emission transition is emitting UV light, will tell you that this is a part of the Lyman series, and thus the transition is going from a higher energy level down to n=1.

This is important because this tells you that the final energy level of this transition is n=1. You can not solve the problem without knowing this information.

I would highly recommend memorizing these hydrogen series for the midterm

1. Lyman: UV light emission, with the transition ending in n=1

2. Balmer: visible light emission, with transition ending in n=2

3. Paschen: IR light emission, with transition ending in n=3

It will also be helpful to memorize the general trends of the electromagnetic spectrum

[Ashwin Vasudevan 3A]

Hello,
You are using the right equation for this question. To start I would find "v" or the frequency. Using the wavelength = c/v you can plug in wavelength and speed of light to figure out the frequency. Using the frequency and n1=1 from the balmer series rule, you can solve for n2.

[205677523]

You are correct! The bit of information you're missing is that a 102.6 nm emission is UV radiation, suggesting that this emission is part of the Lyman series, meaning that the final energy level the electron ends up in is n = 1. Plug that in for n2 and solve for n1!

Hello, when I put n=1 as n2 I got a negative number, so I assume then I should use n=1 as my n1 value, but how do I know whether to do that in future questions?

[Jacqueline Musico 3A]

^@205677523 Yes, you are correct that n=1 should be the n1 value. The book expresses the Rydberg equation where n1 = 1,2,3... (discrete values) and n2 = n1+1, n1+2, n1+3... For the Balmer series, n1 is always equal to 2, and in the Lyman series, n1 is always equal to 1. n1 is the smaller number because it represents the lower energy level that the electron transitions to (its final state).

[Julie Tran 3C]

You are correct! The bit of information you're missing is that a 102.6 nm emission is UV radiation, suggesting that this emission is part of the Lyman series, meaning that the final energy level the electron ends up in is n = 1. Plug that in for n2 and solve for n1!

Hello, when I put n=1 as n2 I got a negative number, so I assume then I should use n=1 as my n1 value, but how do I know whether to do that in future questions?

In general, the first term (minuend) should be the final energy level.