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sasquatch

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Ok there a question that i dont get:

5. If the atoms in a sample of hydrogen were in the state n = 5, how many differnent spectral lines could be produced by the gas as the electrons returned to the ground state.

I was confused with this question and my girlfriend pointed out that hydrogen only has one electron in its shell so thats a weird thing. Even if it didnt have one electron, wouldnt it moving from one state to another release a photon which has a frequency correspoinding to the amount of energy released? so yeah wouldnt it only be 1 spectral line? The answer is 10...

Also with these questions:

Given that E1 = - 13.6 eV, E1 = -3.40 eV, E3 = -1.51 eV

Calculate the wavelength of the first two lines in the Lyman series (dropping from higher state to first state)

So the photon released when dropping from a higher state to a lower state.. shouldnt it be the /\E = final amount of energy - initial amount of energy,

which means /\E = -13.6 -(-3.4) = -10.2 eV

But that is not right cuz the photon energy E = hf = -10.2 does not make sense as we know h = 6.63x10^-34 and f has to be positive, so why is it actually

/\E = -3.4 -(-13.6) = 10.2 (or initial - final) cuz thats confusing.. i know it has to be positive so i can assure myself that way.. but could anybody give me a real explination.

Thanks!
 

aspraveen

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Can I please know where I can get some latest trial CSSA trials form(2005)
 

angmor

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E1 = - 13.6 eV, E1 = -3.40 eV, ?

do u mean that E2 = -3.4 eV?
 

sasquatch

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umm nuh i dunno.. i dont have it nor do i know where to get it from..
 

angmor

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ok so E1 = - 13.6 eV, E2 = -3.40 eV, E3 = -1.51 eV

i think you have to use the Balmer equation which is

1/wavelength = R[ (1/final state^2) - (1/initial state^2) ]

so say you wanted to calculate the wavelength from E2 to E1, you just use E2 as the initial state, and E1 as the final state dont you? correct me if im wrong im not too sure :)
 

sasquatch

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Yeah i know you can do it that way...

but you can also do it the way:

hf = energy lost by moving to a lower state

f = c/wavelength

so yeah but my question wasnt how you do it it was, why is the
Energy lost = initial amount of energy - final amount of energy

or whatever
 

angmor

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That is because when you say energy lost, what you also mean is the change in the amount of energy, i.e. delta E.

Now E=hf. The higher you go up the spectral lines, the higher the frequency.
Thus, as h is a constant, the higher your frequency, the higher your amount of energy.
Therefore the initial amount of energy has a higher amount of energy, since it has a higher state. So
delta E = Ei - Ef

Hope that helps :)
 

Riviet

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sasquatch said:
Ok there a question that i dont get:

5. If the atoms in a sample of hydrogen were in the state n = 5, how many differnent spectral lines could be produced by the gas as the electrons returned to the ground state.

I was confused with this question and my girlfriend pointed out that hydrogen only has one electron in its shell so thats a weird thing. Even if it didnt have one electron, wouldnt it moving from one state to another release a photon which has a frequency correspoinding to the amount of energy released? so yeah wouldnt it only be 1 spectral line? The answer is 10...
Remembering that the ground state is when n=1, the number of different spectral lines can be found by counting the number of ways the electrons can jump from one energy level to the previous:
From the 5th, the electrons can drop to the 4th or 3rd or 2nd or 1st, which gives 4 ways to start their return.
If the electrons have returned to the 4th, from here they can go down to the 3rd or 2nd or 1st energy level, giving 3 more paths to take.
You should be able to see that if the electrons drop down to the 3rd level, then there are two left to drop to and after dropping down to the 2nd level, it can only drop 1 more to reach the ground state.
So adding up all the possible paths that the electrons can take:
number of spectral lines = no. of paths from 5th energy state + no. of paths from 4th energy state + no. of paths from 3rd energy state + no. of paths from 2nd energy state
=4 + 3 + 2 + 1
=10

P.S I just did this question from the Jacaranda textbook today. :)
 
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alcalder

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And yes, there is only 1 electron in a hydrogen atom, but there are many hydrogen atoms in a gas of hydrogen. Therefore, Every single one of those atoms could be dropping its electron to the ground state through different paths.

Thus, multiple lines are produced.

The question does not ask this, but also you could think, how many of those spectral lines lie within the visible light spectrum? Only the Balmer series produces visible light spectral lines (and that is drop to n=2). If n=5 is the upper limit of your outer shell, then only transitions n=5 to 2, n=4 to 2 and n=3 to 2, will produce visible line spectra. Overall, however, you will have 10 lines, some in the infra red (dropping to n=4 and 3), some in the ultra violet (dropping to nb1) as well.

Does that help?
 

sasquatch

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And yes, there is only 1 electron in a hydrogen atom, but there are many hydrogen atoms in a gas of hydrogen. Therefore, Every single one of those atoms could be dropping its electron to the ground state through different paths.
Yeah i didnt read the question properly.

But from Riviet's explination i get it now, thanks everyone for their help.
 

aus_peter2005

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i think the negative sign for the change in energy (ie delta E) mean that the atom has lost some energy and becomes more stable??????....i am not sure if this is right....that's what i learn from chemistry...when calculating value for frequency i think you simply take the absolute value of the delta E....since the sign for the delta E represents the direction of energy flow........hope that helps~~~~~
 

sasquatch

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oh well i dunno.. but ill just take the positive answer anyway..hehe
 

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