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Ionisation Energy of Hydrogen

  • Thread starter mrausum
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Homework Statement



The wavelengths λ of spectral lines produced by the Hydrogen atom are given by the expression:

7c53df3d1d6bba9ca9ba2a89f7f3baae.png
.

Calculate the Energy in eV required to ionise the atom

Homework Equations



p=h/[tex]\lambda[/tex]

The Attempt at a Solution



n2 = 1 and n1 = infinity, therefore [tex]\lambda[/tex]=1/R.

E = p2/2m = h2/([tex]\lambda[/tex]^2*2Me)

Subbing in [tex]\lambda[/tex]=1/R. and rearranging gives:

E = (h2R2)/(2Me)

Why doesn't this give me the right answer?
 
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Answers and Replies

  • #2
Redbelly98
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Homework Equations



p=h/[tex]\lambda[/tex]
In that equation, λ is the deBroglie wavelength of the electron, and not the wavelength of the photon. So I'm afraid it isn't relevant here.

Instead, what other equation do you have that relates the energy and wavelength of a photon?
 
  • #3
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In that equation, λ is the deBroglie wavelength of the electron, and not the wavelength of the photon. So I'm afraid it isn't relevant here.

Instead, what other equation do you have that relates the energy and wavelength of a photon?
So it's just as simple as using E = hc/lamda = hcR?
 
  • #4
Redbelly98
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Yes. :smile:
 
  • #5
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ah i see what's going on now - the equation gives the wavelength of the emitted photon not the wavelength of the electron? Thanks.
 
  • #6
Redbelly98
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ah i see what's going on now - the equation gives the wavelength of the emitted photon not the wavelength of the electron?
Yup!

Thanks.
You're welcome.
 

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