Termoelectric Emission: Measuring T & I_A for Work

[Petar Mali]

I_A=BT^2e^{-\frac{e\varphi}{k_BT}}

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

\frac{I_A}{I_{A0}}=\frac{T^2}{T^2_0}e^{-\frac{e\varphi}{k_BT}(\frac{1}{T}-\frac{1}{T_0})}

And from that we get

\varphi=-\frac{k_B}{e(\frac{1}{T}-\frac{1}{T_0})}ln\frac{I_AT^2_0}{I_{A0}T^2}

and from that they calculate tangent like

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

What happened with T_0?

 

[ZapperZ]

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

Zz.

 

[Petar Mali]

OK! You don't have any more \frac{1}{T}-\frac{1}{T_0}. In last equation you have \frac{1}{T} instead.

 

[Dickfore]

lol.what does \Delta\frac{1}{T} stand for? And T_{0} occurs once more in the equation. Can you find it?

EDIT:

I think the numerator should have \Delta \ln(\frac{I}{T^{2}}) instead of what you wrote, because you are looking for the slope. Put differently, if you plot \ln(\frac{I}{T^{2}}) vs. \frac{1}{T} (for several T's and not just two), then you should get a line with a slope proportional to the work function.

 

[Petar Mali]

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


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


to get that work. THAT IS IN SCRIPTS. WHAT HAPPENED WITH -\frac{1}{T_0}, If that is easier for you?

 

[Dickfore]

It is one of your measurements!

 

[Petar Mali]

Can you understand that. Thank you!