1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Urgent: de Brogile wavelengths of atoms, working in eV!

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data
    http://www.strings.ph.qmul.ac.uk/~russo/QP/r06QPHYEX.pdf [Broken]
    Question 3.

    3. The attempt at a solution
    I can easily derive the equation I need, which is:

    lamda = h/p

    Which after some playing around with K.E =1/2mv^2 etc we obtain:

    lambda (de brogile) = h/SQRT(2mKE)

    Here is the data we are provided with in the exam (see sheet 1): http://www.strings.ph.qmul.ac.uk/~russo/QP/week6.pdf [Broken]

    Everythings in eV and MeV, I plug in the values constantly, tried converting h to joules, and the electron mass to kg etc, but no avail as I don't think I know what Im doing with it, and I can't seem to obtain the real answer which is: 1.2 x10^-9m

    Anyone explain how to do this quickly? I understand the theory behind this top notch but missed on how to use eV etc.

    THANKS.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 13, 2009 #2

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    The equation appears to be right (I plugged in values and got the right answer), what kind of answers are you getting? It's probably just an arithmetic error, or a unit error.
     
  4. May 13, 2009 #3
    Thanks for the reply!

    I plugged in:
    4.14x10^-21 / SQRT(2 x 0.511 x 1.6x10-24) = 3.2 x 10-9.

    I converted 1eV into MeV. Where am I going wrong with this?
    If I convert everything to eV and use:
    4.14x10^-15 / SQRT(2 x 511 x 1.6x10-19) = 3.2 x 10-7m

    Still wrong, and I'm ripping my hair out on this!
     
  5. May 13, 2009 #4

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    Well, when you use electron volts, what speed comes out? It's not meters/second, but rather c.

    Also, you have some wrong values. The electron is 511keV = .511MeV (correct for your calculation in MeV) = 511000eV (incorrect when you plugged it in eV).

    Also, you're converting 1 eV into joules which doesn't make sense since you used everything else in eV.
     
  6. May 13, 2009 #5
    So how would I go about doing this?

    Would I multiply the value of the rest mass by (3x10^8)^2 to obtain the MeV value?
     
  7. May 13, 2009 #6

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    no, because then you get a dimensionless unit (or /s or something weird like that).

    What you want to realize is first, don't convert your 1eV into 1.602*10^-19 Joules, and 2 at the end realize that your answer is given as a fraction of c (times seconds).

    How would you then convert the answer to the correct one? Well, multiply your answer by the value of c.

    The units will be weird, but the procedure should get you the right answer at the end.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Urgent: de Brogile wavelengths of atoms, working in eV!
  1. De broglie wavelength? (Replies: 1)

  2. De broglie wavelength (Replies: 2)

Loading...