Minimum wavelength when electrons strike phospor

  • Thread starter songoku
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  • #1
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Homework Statement


Assume electrons are accelerated through a potential difference of 20000 V inside a TV picture tube. What is the minimum wavelength that could be produced when the electrons strike the phosphor?


Homework Equations


W = qV



The Attempt at a Solution


I use the equation above to find the speed, then plug it in de Broglie formula to find the wavelength but there is no such answer in the multiple choices....Maybe we need to know the work function of phospor?

Thanks
 

Answers and Replies

  • #2
kuruman
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To get help, please show your work and the possible answers. Otherwise, we can't tell what you did wrong, if anything.
 
  • #3
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Hi kuruman

Possible answers :
a. 0.62 angstrom
b. 1.15 angstrom
c. 10.5 angstrom
d. 100.9 angstrom

My work :

[tex]W=qV[/tex]

[tex]\frac{1}{2}mv^2=qV[/tex]

[tex]v=8386276.94~ms^-1[/tex]

Then

[tex]\lambda = \frac{h}{mv}[/tex]

[tex]=8.68\times 10^-12[/tex]

[tex]=0.087~\text{angstrom}[/tex]
 
  • #4
kuruman
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What you calculated here is the wavelength of a 20 keV electron, not the photon that it produces. What is the relation between energy and wavelength for a photon? You can use the de Broglie expression, but momentum is not mv for a photon. What is it?
 
  • #5
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Hi kuruman

Momentum for a photon = E/c , where E = hf

Then, the equation will be :

[tex]p=\frac{E}{c}=\frac{hf}{c}=\frac{h}{\lambda}[/tex]

[tex]\lambda = \frac{h}{p}[/tex]

I don't know how to find the momentum.....

Thanks
 
  • #6
kuruman
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You don't really need the momentum. You know that the energy of the photon is 20 keV and you are looking for its wavelength. You already have the expression

E/c = h/λ

so go for it.
 
  • #7
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Hi kuruman

Oh now I get it. Thanks a lot ! :smile:
 

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