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De Broglie Wavelength Calculation given electron velocity

  • Thread starter msc8127
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  • #1
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Homework Statement


an electron is moving at 2.5 x 105m/s. Find the electron's De Broglie Wavelength.


Homework Equations



De Broglie relates wavelength of electron waves to momentum of the electron by lambda = h/p where h is planck's constant.


The Attempt at a Solution



given the velocity above of 2.5 x 105m/s, I multiplied that velocity by 9.109 x 10-31kg to get momentum (p).

from there i used (4.136 x 10-15eVs) / (9.109 x 10-31kg)(2.5 X 105m/s)

This gives me 18162257108.4 m which is off by many orders of magnitude from the 180pm that the book lists as the answer.

Can someone please show me where i'm going stupid on this one?

Thanks!
 

Answers and Replies

  • #2
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Homework Statement


an electron is moving at 2.5 x 105m/s. Find the electron's De Broglie Wavelength.


Homework Equations



De Broglie relates wavelength of electron waves to momentum of the electron by lambda = h/p where h is planck's constant.


The Attempt at a Solution



given the velocity above of 2.5 x 105m/s, I multiplied that velocity by 9.109 x 10-31kg to get momentum (p).

from there i used (4.136 x 10-15eVs) / (9.109 x 10-31kg)(2.5 X 105m/s)

This gives me 18162257108.4 m which is off by many orders of magnitude from the 180pm that the book lists as the answer.

Can someone please show me where i'm going stupid on this one?

Thanks!
? That's not planck's constant.
 
  • #3
23
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6.626 x 10^-34 Js is actually planck's constant. I tried that and I get the correct order of magnitude, but instead of getting 180pm i get 290pm....the number posted in the original post is plancks constant converted to eV (which i'm sure you already knew)

Either form of planck's constant that i use I'm getting an incorrect solution

thank you
 
  • #4
Redbelly98
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The Attempt at a Solution



given the velocity above of 2.5 x 105m/s, I multiplied that velocity by 9.109 x 10-31kg to get momentum (p).

from there i used (4.136 x 10-15eVs) / (9.109 x 10-31kg)(2.5 X 105m/s)

This gives me 18162257108.4 m ...
No, that's wrong because the units do not work out to be meters. Your answer here should be 18162257108.4 eV*s2/kg*m.

... which is off by many orders of magnitude from the 180pm that the book lists as the answer.
Argh, the book is wrong. Your answer of 290 pm is better, you just have the wrong power of 10 in your calculation.

Note for future: when dealing in kg, m, and s then use J for energy, not eV.
 
  • #5
I'm having another problem too. It's not the same but I have been looking everywhere to find my answer and I don't know if I am correct. I used this walkthrough I found to get my answer but I used a different wavelength for my experiment. Here's the link.
http://wiki.answers.com/Q/What_is_the_speed_of_the_electron_with_a_de_Broglie_wavelength_of_235nm

I did the same thing except I used a wavelength of 400 nm and my answer was 2038 m/s. This is one of my first problems with this and I was just curious if anyone could comfirn this answer or if I failed horribly.

Thanks.
 
  • #6
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
I'm having another problem too. It's not the same but I have been looking everywhere to find my answer and I don't know if I am correct. I used this walkthrough I found to get my answer but I used a different wavelength for my experiment. Here's the link.
http://wiki.answers.com/Q/What_is_the_speed_of_the_electron_with_a_de_Broglie_wavelength_of_235nm

I did the same thing except I used a wavelength of 400 nm and my answer was 2038 m/s. This is one of my first problems with this and I was just curious if anyone could comfirn this answer or if I failed horribly.

Thanks.
You're a little off but not horribly off.

For future reference, please start a new thread with your question, even if it is related (but not identical) to an existing one.
 
  • #7
Thanks I'll keep that in mind!! I just joined this forum so yeah.

Thanks for your help!
 

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