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ChemSiddiqui
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Hey everyone,

 

I was going through some past exam papers and this question came up and I am as the title says stuck. here goes the problem;

 

what states of hydrogen atom have an ionisation energy of 5.448 x 10^(-19) J?

 

This question requires students to do some mathematical calculations. I was wondering if can I use the formulea

[math]

Enx-Eny = hcR (\frac{1}{nx^2} - \frac {1}{ny^2})

[/math] and then apply [math] En = - \frac {hcR}{n^2}[/math] (where R is the rydberg constant) to get n and work out the quantum numbers from there?

 

any help most appreciated. thanks

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Isn't the ionization energy simply the magnitude of the binding energy? The binding energy goes with the so-called principal quantum number n as something/n² (probably like your 2nd equation). So yes, you can get n from that and then work out all states belonging to that n.

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do you mean [math] En = -R \frac{Z^2}{n^2} [/math]? but then the question gives a positive value of ionisation energy and with this expression we cant get the value of n as the underoot will be complex. I think if we used the first equation in my previous post such that to set ny = infinity and then from there we can work out nx? what you say to that? thanks for your help though!

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With "magnitude of" I meant the absolute value of the binding energy, a positive value. Seems like I picked the wrong word.

 

You can formally use the other equation and set one of the ns to infinity. That will give you the same result except that it is harder to write down correctly (how do you divide something by infinity? Is infinity even a sensible natural number? ...) and seems completely unmotivated to me.

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hmm....Ok!. I just did the calculation and the result seems a little absurd. Here goes;

 

[math] E= 5.448 X 10^-(19) J; n=?; R=1.0967 X 10^7 m^-1; Z=1 for H atom[/math]

 

[math] n^2 = \frac{RZ^2}{E} => n = 4.48 X 10^12. [/math]

 

what have I done wrong here? can you let me know? Thanks

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n=4 makes more sense but I still doubt it. In fact, I have a good idea what went wrong this time so just go through your calculation step by step again. I had my username changed by asking an admin in the chat.

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