Bohr Quantisation Question

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Homework Statement


The ionisation energy of a single hydrogen atom is 13.6 eV. The application of Bohr quantisation to the hydrogen atom results in the stationary states having discrete energies given in terms of a positive integer n according to

E=-RB/n2

where RB is the Rydberg constant. Determine the value of the Rydberg constant. Assuming like Bohr that, when an atom emits or absorbs radiation, it does so in the form of a single quantum, compute the wavelength of the spectral line of the Balmer series (the transition from n=3 to n=2) for hydrogen.

Homework Equations


E=-RB/n2
[tex]\stackrel{1}{\lambda}[/tex]=RB[[tex]\stackrel{1}{n12}[/tex]-[tex]\stackrel{1}{n22}[/tex]]


The Attempt at a Solution


Ok my main issue with this whole question is which value is the Rydberg constant. Is it:
2.18*10-18J or 1.097373157*107m-1 (or both)

If I go with the first value, I can easily obtain that using the first equation as it mentions in the equation "it does so in the form of a single quantum", meaning that n in the equation is equal to 1 and its a simple case of multiplying 13.6 eV by 1.6*10-19 to convert it to joules giving 2.18*10-18J.

For the second part however I cannot get a reasonable answer using this value of the Rydberg constant, only with the second one as shown below:

Using first value:
[tex]\stackrel{1}{\lambda}[/tex]=(2.18*10-18)[[tex]\stackrel{1}{4}[/tex]-[tex]\stackrel{1}{9}[/tex]]
lambda=3.03*10-19m

Using second value:
[tex]\stackrel{1}{\lambda}[/tex]=(1.097373157*107)[[tex]\stackrel{1}{4}[/tex]-[tex]\stackrel{1}{9}[/tex]]
=656nm (visible)

Im guessing that the constant is both values but I've missed a conversion somewhere. Either that or there is something missing in the second equation to amend this issue with using 2.18*10-18J.

Finally, how do you do fractions on this properly because as you can see on my post, its seperated each fraction from the equation and has somehow enlarged itself. I havent used this forum that much so i'm still unsure on things like this.

Thanks in advance
 

Answers and Replies

  • #2
vela
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There's the Rydberg constant, which you found correctly to be

[tex]R_\infty = 1.097\times10^7~\textrm{m}^{-1}[/tex]

and the related Rydberg unit of energy, which is

[tex]1~\textrm{Ry} = hcR_\infty = 13.6~\textrm{eV}[/tex]

Your first equation uses the energy; your second equation uses the Rydberg constant.

To write the equation

[tex]\frac{1}{\lambda} = R_B\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)[/tex]

you'd use the code

Code:
\frac{1}{\lambda} = R_B\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

sandwiched between the tex tags. You should also be able to right-click on the equation to see the TeX used, though I couldn't get it to work here just now.
 
Last edited:
  • #3
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Thanks for explaining clearly which equation is for which. I just now went through converting the energy to the rydberg constant (per m) and it turns out it was alot easier than I initially thought. Also im very grateful for the code you've displayed regarding fractions on here. It turns out I actually used the wrong symbol in the latex reference, but I've found the real fraction symbol now so it's all good.
 

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