Hi for part c given that there are some repeats in the difference between 2 squares, e.g. 1&49 and 16&64, the number of observable lines would be less than 45 right? I counted and there are 41 distinct differences of 2 squares for n=1 to n=10
I think not actually; we’re interested in the difference between one over the squares, so things like 1/3^2 – 1/5^2 . There’s 45 of these because these should all be different.
Hi, question c is confusing me, wouldn’t there be more than 45 as they don’t necessarily jump from 10 to 9 to 8 but could also jump from 10 to 7 to 4 etc, releasing more wavelengths?
Your reasoning is absolutely right: you could have jumps from 10 to any of the levels 1 – 9, from 9 to any of 1-8, and so on (the only constraint is that you have to go from high to low if you are going to emit any light). But the total number of these pairs is 45.
Thanks for replying, I understand where the wavelengths are coming from now, I was counting each difference of 1 as a separate wavelength instead of each difference being separate if that makes any sense.
Hi, I was wondering for part B, how do we know that the change in energy levels should be from n=10 to n=9, and not, say, n=2 to n=1? Should we assume that the initial energy level of the atom is always n=10? Thanks so much for your website by the way, it has been incredibly helpful!
The question in B asks for the longest wavelength for any change from level n to level m. But if you look on the third line of the answer at the equation we have derived for wavelength of light resulting from a level change, you should be able to see that the maximum value of the wavelength comes when we change from 10 to 9.
In part C however, we are considering all possible wavelengths, which means we consider all values of n and m from 1 to 10 inclusive.
I think for question C the answer should be 7 because only when the level falls to level 2 will produce observable light, which means only 10 to 3 can produce observable light. Thus the answer should be 7
Well, yes and no. Yes, as it turns out the only emissions of hydrogen visible to a human eye are when the lower energy level is 2. But no, physics the word “observed” does not just mean “observed by the naked eye” — it really means “detected (using any suitable apparatus)”. So, while most of the lines are in the UV or IR parts of the spectrum, they can still be detected using some sensors and so they can all be observed.
In the exam, unless a question specifically refers to the human senses in some way, you should assume that ‘observable” means “could be detected by some sensor system”
At energy level 10, we can generate 9 emission lines, corresponding to the photons emitted when we drop from 10 to 9, 10 to 8, 10 to 7, and so on down to 10 to 1. At energy level 9 we can generate 8 emission lines, corresponding to the photons emitted when we drop from 9 to 8, 9 to 7, 9 to 6, and so on down to 9 to 1.
In general, at energy level n we generate (n-1) emission lines. So the total number of emission lines is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, corresponding to the lines generated by energy levels 2, 3, 4, 5, 6, 7, 8 and 9 respectively.
The notation Σn, where 1 ≤ n ≤ 9, just means 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9.
Alternatively, one could also use the binomial coefficient (n choose k) here to determine the number of possible different combinations of two energy levels, which are just the same as the number of emission lines. In this example it would be (10 choose 2) which is 10!/(2!(10-2)!) = 109/2 = 45.
Now that I think about it, the sum formula (x(x+1)/2) is actually just a simplified version of the binomial coefficient formula for k = 2, which I think is pretty interesting. So, if one somehow happens to only know the formula for binomial coefficients and not the one for sums, this might be a neat shortcut to calculating large sums of this kind by hand.
Hi, this topic isn’t really going to be covered for my A Levels until November, so do you have any useful links where I can learn about quantum mechanics(if this is that topic)?
Thankyou.
To answer this question you need to know that the energy E of a photon is related to its frequency f by the equation E = hf where h is the Planck constant that the question refers to. The only bit of the syllabus that covers this is “atomic structure”, which is a bit cryptic. You could try looking at this site: http://www.bbc.co.uk/education/topics/znv39j6, especially the sections on wave particle duality and spectra.
Oxford PAT 
Hi for part c given that there are some repeats in the difference between 2 squares, e.g. 1&49 and 16&64, the number of observable lines would be less than 45 right? I counted and there are 41 distinct differences of 2 squares for n=1 to n=10
Hi Kei,
I think not actually; we’re interested in the difference between one over the squares, so things like 1/3^2 – 1/5^2 . There’s 45 of these because these should all be different.
Hi, question c is confusing me, wouldn’t there be more than 45 as they don’t necessarily jump from 10 to 9 to 8 but could also jump from 10 to 7 to 4 etc, releasing more wavelengths?
Your reasoning is absolutely right: you could have jumps from 10 to any of the levels 1 – 9, from 9 to any of 1-8, and so on (the only constraint is that you have to go from high to low if you are going to emit any light). But the total number of these pairs is 45.
Thanks for replying, I understand where the wavelengths are coming from now, I was counting each difference of 1 as a separate wavelength instead of each difference being separate if that makes any sense.
Wait no I get it never mind
Wait no ignore that, I’ve got it
Ignore that last comment, I understand it now
Hi, I was wondering for part B, how do we know that the change in energy levels should be from n=10 to n=9, and not, say, n=2 to n=1? Should we assume that the initial energy level of the atom is always n=10? Thanks so much for your website by the way, it has been incredibly helpful!
The question in B asks for the longest wavelength for any change from level n to level m. But if you look on the third line of the answer at the equation we have derived for wavelength of light resulting from a level change, you should be able to see that the maximum value of the wavelength comes when we change from 10 to 9.
In part C however, we are considering all possible wavelengths, which means we consider all values of n and m from 1 to 10 inclusive.
Alright, thank you for the clarification!
A very interesting and beautiful question. Parts of the learning material for this question are not only in “Natural world” but also in waves, right?
Yes that’s right. But you only need to know that wavelength = wave speed / frequency.
I think for question C the answer should be 7 because only when the level falls to level 2 will produce observable light, which means only 10 to 3 can produce observable light. Thus the answer should be 7
Well, yes and no. Yes, as it turns out the only emissions of hydrogen visible to a human eye are when the lower energy level is 2. But no, physics the word “observed” does not just mean “observed by the naked eye” — it really means “detected (using any suitable apparatus)”. So, while most of the lines are in the UV or IR parts of the spectrum, they can still be detected using some sensors and so they can all be observed.
In the exam, unless a question specifically refers to the human senses in some way, you should assume that ‘observable” means “could be detected by some sensor system”
Thanks so much for the website it’s really useful!
Please could you explain your answer to part c? Sigma notations don’t come naturally to me
At energy level 10, we can generate 9 emission lines, corresponding to the photons emitted when we drop from 10 to 9, 10 to 8, 10 to 7, and so on down to 10 to 1. At energy level 9 we can generate 8 emission lines, corresponding to the photons emitted when we drop from 9 to 8, 9 to 7, 9 to 6, and so on down to 9 to 1.
In general, at energy level n we generate (n-1) emission lines. So the total number of emission lines is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, corresponding to the lines generated by energy levels 2, 3, 4, 5, 6, 7, 8 and 9 respectively.
The notation Σn, where 1 ≤ n ≤ 9, just means 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9.
Alternatively, one could also use the binomial coefficient (n choose k) here to determine the number of possible different combinations of two energy levels, which are just the same as the number of emission lines. In this example it would be (10 choose 2) which is 10!/(2!(10-2)!) = 109/2 = 45.
Now that I think about it, the sum formula (x(x+1)/2) is actually just a simplified version of the binomial coefficient formula for k = 2, which I think is pretty interesting. So, if one somehow happens to only know the formula for binomial coefficients and not the one for sums, this might be a neat shortcut to calculating large sums of this kind by hand.
Yes you’re right that would be a nice way of doing it.
Hi, this topic isn’t really going to be covered for my A Levels until November, so do you have any useful links where I can learn about quantum mechanics(if this is that topic)?
Thankyou.
To answer this question you need to know that the energy E of a photon is related to its frequency f by the equation E = hf where h is the Planck constant that the question refers to. The only bit of the syllabus that covers this is “atomic structure”, which is a bit cryptic. You could try looking at this site: http://www.bbc.co.uk/education/topics/znv39j6, especially the sections on wave particle duality and spectra.