Identify emission line spectrum

  • #1

Homework Statement


refer to the question

Homework Equations




The Attempt at a Solution


from figure b , moving to the right, we know that the wavelength increases. since e= hc/λ , so the energy diffrenece should be increases as going to the right, so i consider line 1,2 and 3 is in a specific series , and 4,5 in another series. line 1,2,3 should have higher energy difference , the question ask for electron transition ftom E1 to E0 . So , my ans is line 3 . but the line 3 is not choice for this question. the ans is line 5 . why? [/B]
 

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Answers and Replies

  • #2
BvU
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Hi Des. Posting goes against a whole lot of PF guidelines -- as you probably are well aware.
So I can't state here that I agree with your reasoning and the book answer must be in error.
 
  • #3
NascentOxygen
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, the question ask for electron transition ftom E1 to E0 . So , my ans is line 3 . but the line 3 is not choice for this question. the ans is line 5 . why?
I would have said line 3, too.
 
  • #4
rude man
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Hard to say. There are 6 possibilities, not 5, but E1 to E0 and E3 to E1 are about the same. I guess both of these could be 3.
 
  • #5
andrevdh
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Yes the wavelength increases to the right, but the energy of the produced photons decreases to the right according to your formula since the energy is inversely proportional to the wavelength, that is the energy of the photons decreases as the wavelength increases. So the higher energy lines is on the left and the lower energy on the right. The key to the answer is in the remark rude man made. The wavelengths and energies of the two transitions he mentions are close together. Why? And which transition will give the shorter wavelength of the two?
 
  • #6
NascentOxygen
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We are told it's the hydrogen spectrum. So we know the Lyman Series is well-separated from the Balmer Series. This would seem to leave no room for debate.
 
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  • #7
ehild
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The energy scale in the figure can be logarithmic. :)

ehild
 
  • #9
ehild
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so can someone draw a conclusion?

My conclusion is that the official answer is wrong as it happens very frequently.
 
  • #12
BvU
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It's physics, not some kind of referendum, for goodness sake. You've had four heavyweights backing you up !
 
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  • #13
NascentOxygen
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so can someone draw a conclusion?
Was there a 6th line in the spectrum, one which got blotted out? Just wondering....

When Q doesn't match A, sometimes it's the question which is wrong. Not always is the textbook's answer at fault :w
 
  • #14
BvU
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We are told it's the hydrogen spectrum
No we are told these are some of the energy levels in the H atom.

It could well be the student only has the 1/n2 expression for the energy levels at this point in the curriculum, (he doesn't give any relevant equations at all) and in that case, from the ratios of the energy differences (assuming a linear scale, I admit), E0 in the picture is for n = 4 with reasonable certainty. Gives wavelengths in the micrometer range (2.2, 2.6, 4.1, 4.7, 7.5, 12.4). So even then, the wavelength picture doesn't fit qualitatively.

But for sure E1 -> E0 is the center of the five lines (where we still have to assume the sixth is off-scale to the right -- or cut off by the student :-) ).

Too many assumptions, pictures too vague. What book is this from ?
 
  • #16
NascentOxygen
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No we are told these are some of the energy levels in the H atom.
We are clearly told the line spectrum pertains to the hydrogen atom.
 
  • #17
BvU
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Why is it so? Thank you all mentor!!
Why is what so ?

Homework Statement


Figure (a) below shows some of the energy levels of a hydrogen atom ....

is what the text says. We are somehat brainwashed to think E0 = -13.6 eV, but the author can name the levels any way he wants -- if he is devious enough. I had fun measuring the y coordinates on my screen and found 0, 79, 124, 149 mm. Matches 0, 1/16-1/25, 1/16-1/35, 1/16-1/49 suspiciously well. in eV that would give 0, 0.306, 0.472, 0.572.

With

Homework Equations


$$ E = {hc \over \lambda} $$
So ##\lambda = {hc\over eE}## if E is in eV (e = 1.6e-19 Coulomb), the wavelengths I came up with -- and that don't match the righthand Figure (b) :(
 

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