Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Termoelectric emission

  1. May 18, 2010 #1

    We measure [tex]T[/tex] and [tex]I_A[/tex] and from that get work. If [tex]T_0[/tex] is first temperature in which we measured than
    we get


    And from that we get


    and from that they calculate tangent like

    [tex]\varphi=-\frac{k_B}{e}\frac{\Delta ln\frac{I_AT^2_0}{I_{A0}T^2}}{\Delta \frac{1}{T}}}[/tex]

    What happened with [tex]T_0[/tex]?
  2. jcsd
  3. May 18, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Your question is puzzling. What is it exactly that you didn't understand? [itex]T_0[/itex] is still in all your equations. As far as I can see, nothing happened to it.

  4. May 18, 2010 #3
    OK! You don't have any more [tex]\frac{1}{T}-\frac{1}{T_0}[/tex]. In last equation you have [tex]\frac{1}{T}[/tex] instead.
  5. May 18, 2010 #4
    lol.what does [itex]\Delta\frac{1}{T}[/itex] stand for? And [itex]T_{0}[/itex] occurs once more in the equation. Can you find it?


    I think the numerator should have [itex]\Delta \ln(\frac{I}{T^{2}})[/itex] instead of what you wrote, because you are looking for the slope. Put differently, if you plot [itex]\ln(\frac{I}{T^{2}})[/itex] vs. [itex]\frac{1}{T}[/itex] (for several T's and not just two), then you should get a line with a slope proportional to the work function.
    Last edited: May 18, 2010
  6. May 18, 2010 #5
    Yes I work that for several T and get graph. From that graph I must find slope. But please look at lines that I wrote. My problem is because I use

    [tex]\varphi=-\frac{k_B}{e}\frac{\Delta ln\frac{I_AT^2_0}{I_{A0}T^2}}{\Delta \frac{1}{T}}}[/tex]

    to get that work. THAT IS IN SCRIPTS. WHAT HAPPENED WITH [tex]-\frac{1}{T_0}[/tex], If that is easier for you?
  7. May 18, 2010 #6
    It is one of your measurements!
  8. May 18, 2010 #7
    Can you understand that. Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook