Physics/Chem Question

[Pisiform]

One of the emission lines of Hydrogen atom has the wavelength of 93.8 nm. Determine the initial and the final values of 'n' associated with this emission.

Ans: n =6; n=1

I first found out Energy of photon by E = hV/(wavelength)

the equate the answer to:

E = -RH [1/n2 - 1/n2 ]

But I don't know how to solve for 'n' in the above equation.

All help is appreciated. Thanks alot . Thank You

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[t1me]

plug+chug approximations in this case would be a lot faster than solving the multivariable equation. Remember, the answers are all given. You just have to pick the right one!

 

[Pisiform]

plug+chug approximations in this case would be a lot faster than solving the multivariable equation. Remember, the answers are all given. You just have to pick the right one!

Cool, I was actually doing some extra practice questions from some other book. The question was not in MCQ form. So thats why i was confused.

 

[ilovemcat]

Edit: Disregard everything I said. I'm confused, lol. An emission spectra would display the wavelength of the light emitted - so there wouldn't be two different wavelengths. Anyways, I really don't know how to do this. 🙂

Cool, I was actually doing some extra practice questions from some other book. The question was not in MCQ form. So thats why i was confused.

Where did this question come from if you don't mind me asking? I stared at it for like 15 minutes, and I still don't understand how you figure out the 'n' values. Did they give 2 wavelengths? Because if they did, then I think you'd be able to solve it like this:

delta E = -2.178x10^-18J / n^2 (some equation for Hydrogen they'd likely provide in a passage), where individual energy levels equal (E=hc/wavelength)

If they gave you two wavelengths (initial and final) then I'd solve it like this:

hc/wavelength #1 = Some Energy Value #1 (E1)
hc/wavelength #2 = Some Energy Value #2 (E2)

delta E = (Efinal/n) - (Einitial /n)and then it's just plug and chug:
E final/delta E = n #2
E initial/delta E = n #1

Sorry if this confused you. This question seems like a total mind f**k. I don't see this being a question on the exam but if anyone knows how to explain or attempt this question, it'd be helpful 🙂

 

[Don Draper]

Edit: Disregard everything I said. I'm confused, lol. An emission spectra would display the wavelength of the light emitted - so there wouldn't be two different wavelengths. Anyways, I really don't know how to do this. 🙂



Where did this question come from if you don't mind me asking? I stared at it for like 15 minutes, and I still don't understand how you figure out the 'n' values. Did they give 2 wavelengths? Because if they did, then I think you'd be able to solve it like this:

delta E = -2.178x10^-18J / n^2 (some equation for Hydrogen they'd likely provide in a passage), where individual energy levels equal (E=hc/wavelength)

If they gave you two wavelengths (initial and final) then I'd solve it like this:

hc/wavelength #1 = Some Energy Value #1 (E1)
hc/wavelength #2 = Some Energy Value #2 (E2)

delta E = (Efinal/n) - (Einitial /n)and then it's just plug and chug:
E final/delta E = n #2
E initial/delta E = n #1

Sorry if this confused you. This question seems like a total mind f**k. I don't see this being a question on the exam but if anyone knows how to explain or attempt this question, it'd be helpful 🙂

don't complicate it.

It is emission, the final n = 1. Now it's just a simple algebra problem solving for the wavelength (or a moderate algebra problem). You can guess n's or solve for the n, either way works. Guessing is probably fastest.

Don't worry, MCAT math is never like this. The hardest MCAT math ever is fractions on top of other fractions, and logs or lots of exponents.

 

[Pisiform]

Edit: Disregard everything I said. I'm confused, lol. An emission spectra would display the wavelength of the light emitted - so there wouldn't be two different wavelengths. Anyways, I really don't know how to do this. 🙂



Where did this question come from if you don't mind me asking? I stared at it for like 15 minutes, and I still don't understand how you figure out the 'n' values. Did they give 2 wavelengths? Because if they did, then I think you'd be able to solve it like this:

delta E = -2.178x10^-18J / n^2 (some equation for Hydrogen they'd likely provide in a passage), where individual energy levels equal (E=hc/wavelength)

If they gave you two wavelengths (initial and final) then I'd solve it like this:

hc/wavelength #1 = Some Energy Value #1 (E1)
hc/wavelength #2 = Some Energy Value #2 (E2)

delta E = (Efinal/n) - (Einitial /n)and then it's just plug and chug:
E final/delta E = n #2
E initial/delta E = n #1

Sorry if this confused you. This question seems like a total mind f**k. I don't see this being a question on the exam but if anyone knows how to explain or attempt this question, it'd be helpful 🙂

Hey bud, they only gave one wavelength and no options (MCQs). I think we have to do it by Trial and Error method. And WHY we need 2 wavelenght? there is only one kind of light is emitted so I believe one wavelenght -- Ok, seriously Idk what I am saying
BTW, this question is from Chemistry - The Central Science Burnsten, Brown

Oh well, I am just moving on because MCAT would have option (answer choices) and in this way it would be much easier.

 

[Don Draper]

Hey bud, they only gave one wavelength and no options (MCQs). I think we have to do it by Trial and Error method. And WHY we need 2 wavelenght? there is only one kind of light is emitted so I believe one wavelenght -- Ok, seriously Idk what I am saying
BTW, this question is from Chemistry - The Central Science Burnsten, Brown

Oh well, I am just moving on because MCAT would have option (answer choices) and in this way it would be much easier.

there is only one wavelength.

The idea is that a certain drop will lead to the emission of a wavelength of light.

While this math is a bit on the harder side for MCAT, the topic isn't. AAMC absolutely LOVES this topic.

 

[ilovemcat]

don't complicate it.

It is emission, the final n = 1. Now it's just a simple algebra problem solving for the wavelength (or a moderate algebra problem). You can guess n's or solve for the n, either way works. Guessing is probably fastest.

Don't worry, MCAT math is never like this. The hardest MCAT math ever is fractions on top of other fractions, and logs or lots of exponents.

Haha, yeah it was late. Can't believe I wrote all that. But still it's a confusing question. I thought I understood this topic well, but I guess I'm wrong. How do you know that the final n = 1? Energy transitions could be from n = 5 -> n = 2 level (Balmer Series), for example. There's no way of telling the final shell value unless they told you - Is there something I'm not realizing here? Lol.

Anyone have a gun I can borrow?

 

[Pisiform]

Haha, yeah it was late. Can't believe I wrote all that. But still it's a confusing question. I thought I understood this topic well, but I guess I'm wrong. How do you know that the final n = 1? Energy transitions could be from n = 5 -> n = 2 level (Balmer Series), for example. There's no way of telling the final shell value unless they told you - Is there something I'm not realizing here? Lol.

Anyone have a gun I can borrow?

Dont worry man, i will lend you a musket 😉

I think n =1 because its hydrogen and it has only one shell . idk seriously

 

[Rabolisk]

There's no reason why final n should be 1. One thing to remember is that the differences in energy between energy levels decrease as you go further away from the nucleus. In this question, though, the given wavelength is clearly in the UV region, which is almost always only possible with the Lyman series (nf=1). The lowest wavelength derived from the balmer series is about 360.

 

[Pisiform]

so do we have to learn that what wavelength fall in Balmer, what fall in Layman, what in Brackett and so on

I didnt learn those 🙁

 

[Rabolisk]

I think you should know that balmer is mainly visible, Lyman is ultraviolet, and anything else will be infrared. At the least it saves time.

 

[ilovemcat]

so do we have to learn that what wavelength fall in Balmer, what fall in Layman, what in Brackett and so on

I didnt learn those 🙁

No, you don't have to know those. I actually never heard of them until I read TBR passages, so they'd definitely explain what they are in a passage.

I noticed you mentioned the final level is n=1 because it's hydrogen. This isn't always true and it's probably important you understand why (because they might have a few questions about this). A Hydrogen electron can absorb energy and jump to a higher energy level. Eventually when it absorbs too much energy, the electron will be ejected off the Hydrogen Atom (Photoelectric Effect). But hypothetically speaking, if a Hydrogen atom absorbs the highest amount of energy possible (without the electron flying off), then it can just as easily emit some of that energy and fall back to the n = 2 level or even the n = 1 level. In each scenario though the change in energy is always the same because Energy Levels are quantized within a specific atom. I hope this didn't confuse you because I'm trying to clarify the concept behind this question which is more likely to show up instead.

 

[Don Draper]

Keep it simple.

No need to know any of these series.

 

[Don Draper]

so do we have to learn that what wavelength fall in Balmer, what fall in Layman, what in Brackett and so on

I didnt learn those 🙁

no.

Just understand that emission is releasing energy and absorption is absorbing, and light is emitted when the energy level drops.

 

[Don Draper]

I think you should know that balmer is mainly visible, Lyman is ultraviolet, and anything else will be infrared. At the least it saves time.

not for the MCAT.

 

[Pisiform]

No, you don't have to know those. I actually never heard of them until I read TBR passages, so they'd definitely explain what they are in a passage.

I noticed you mentioned the final level is n=1 because it's hydrogen. This isn't always true and it's probably important you understand why (because they might have a few questions about this). A Hydrogen electron can absorb energy and jump to a higher energy level. Eventually when it absorbs too much energy, the electron will be ejected off the Hydrogen Atom (Photoelectric Effect). But hypothetically speaking, if a Hydrogen atom absorbs the highest amount of energy possible (without the electron flying off), then it can just as easily emit some of that energy and fall back to the n = 2 level or even the n = 1 level. In each scenario though the change in energy is always the same because Energy Levels are quantized within a specific atom. I hope this didn't confuse you because I'm trying to clarify the concept behind this question which is more likely to show up instead.

I lost you man, literally lost you.
I appreciate it if you explain me once more in an easier way. Thanks 🙂

 

[ilovemcat]

No, you don't have to know those. I actually never heard of them until I read TBR passages, so they'd definitely explain what they are in a passage.

I noticed you mentioned the final level is n=1 because it's hydrogen. This isn't always true and it's probably important you understand why (because they might have a few questions about this). A Hydrogen electron can absorb energy and jump to a higher energy level. Eventually when it absorbs too much energy, the electron will be ejected off the Hydrogen Atom (Photoelectric Effect). But hypothetically speaking, if a Hydrogen atom absorbs the highest amount of energy possible (without the electron flying off), then it can just as easily emit some of that energy and fall back to the n = 2 level or even the n = 1 level. In each scenario though the change in energy is always the same because Energy Levels are quantized within a specific atom. I hope this didn't confuse you because I'm trying to clarify the concept behind this question which is more likely to show up instead.

I lost you man, literally lost you.
I appreciate it if you explain me once more in an easier way. Thanks

This picture is an easier way of explaining what I said earlier. Imagine you have a Hydrogen Atom. The electron is in the n = 1 level, right. Now imagine you input some energy in the form of (UV) light and the Hydrogen Atom absorbs a certain wavelength of light - Let's say 103 nm. Looking at the picture in the diagram below, you see that an absorption of 103 nm wavelength of light corresponds to an energy transition from the n=1 level to n=3.

For a given atom, each energy level has a discrete amount of energy. In the case of Hydrogen:

The n=1 level equals -13.6 eV
The n=3 level equals -1.5ev

So in order for the Hydrogen Electron to jump to the n=3 level, it must absorb the difference of those energies EACH time. In this scenario, the difference in energy (delta E) equals 12.1 eV. Every single time a Hydrogen Atom jumps from the n=1 level to n=3 level, 12.1 eV of energy must be absorbed.

But what I wanted to explain earlier was that energy emission (energy release) doesn't necessarily always drop to the n=1 level. In some cases, it could drop from the n=3 level to the n=2 level. Or it can emit energy from the n=3 level to the n=1 level. Regardless, the energy transitions between two specific energy shells is always the same (as explained above) because each energy level has a set energy value.

If this still confused you, check-out mcat-review.org. They explain things really well.

 

[ilovemcat]

Oh what the heck. It's not showing the pictures I uploaded. grr. Here's the URLs:

http://www.daviddarling.info/images/hydrogen_spectrum.gif
http://www.daviddarling.info/images/hydrogen_atom_energy_levels.jpg

And here's the website they come from:
http://www.daviddarling.info/encyclopedia/H/hydrogen_spectrum.html

 

[Don Draper]

Shot energy at atom?
Absorb and go from n = 1 to n = 2, 3, 4, or whatever

electron falls from n = high # down to n = lower # and light is emitted (emitted energy).

That's pretty much what you need to know.

Quantized means 1,2,3,4,5, etc.

Not: 1, 1.7, 2.54, 3.7, etc

 

[Pisiform]

ok I got it, Thanks a billion Ilovemcat,

But one thing bugging me, a stupid thing (so don't get mad)

its Hydrogen atom and electrons are shifting from n = 2 to n=5 anywhere.
n = number of shell

Don't hydrogen have only 1 shell which has 1 electron. So how other shell come from

Thanks bunch

 

[ilovemcat]

ok I got it, Thanks a billion Ilovemcat,

But one thing bugging me, a stupid thing (so don't get mad)

its Hydrogen atom and electrons are shifting from n = 2 to n=5 anywhere.
n = number of shell

Don't hydrogen have only 1 shell which has 1 electron. So how other shell come from

Thanks bunch

Notice I never said electrons. Like you said, Hydrogen has only one electron. This doesn't mean that 1 electron is confined to the n=1 shell. Only if the Hydrogen Atom is in it's ground state (that is, no energy absorbed) it will remain in the n=1 shell. However, energy absorption allows for the electron to jump to higher energy levels.

Oh and one thing to know - There's an infinite number of shells. n=1 to n=infinity. Theoretically speaking, an electron could occupy any shell (as long as the energy absorbed doesn't exceed its ionization energy). I'm not absolutely sure about this, but from what I remember, Hydrogen is in it's ionized form at about the n=4 level, which is why I didn't include that in the example above.

 

[Pisiform]

Shot energy at atom?
Absorb and go from n = 1 to n = 2, 3, 4, or whatever

electron falls from n = high # down to n = lower # and light is emitted (emitted energy).

That's pretty much what you need to know.

Quantized means 1,2,3,4,5, etc.

Not: 1, 1.7, 2.54, 3.7, etc

Thank you, I hope MCAT only revolve around that

 

[Pisiform]

Notice I never said electrons. Like you said, Hydrogen has only one electron. This doesn't mean that 1 electron is confined to the n=1 shell. Only if the Hydrogen Atom is in it's ground state (that is, no energy absorbed) it will remain in the n=1 shell. However, energy absorption allows for the electron to jump to higher energy levels.

Oh Oh!!!

Man you are AWESOME, hope u get 35 + 👍

It was a stinky Brain Fart

 

[ilovemcat]

Oh Oh!!!

Man you are AWESOME, hope u get 35 + 👍

It was a stinky Brain Fart

It happens, haha. Let's hope we both get 35+ eh? Good luck! 🙂