# Termoelectric emission

1. May 18, 2010

### 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 $$T_0$$?

2. May 18, 2010

### ZapperZ

Staff Emeritus
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.

3. May 18, 2010

### 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.

4. May 18, 2010

### 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.

Last edited: May 18, 2010
5. May 18, 2010

### 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?

6. May 18, 2010

### Dickfore

It is one of your measurements!

7. May 18, 2010

### Petar Mali

Can you understand that. Thank you!