# Electron concentration with temperature change

1. Sep 25, 2009

### snoothie

1. The problem statement, all variables and given/known data

trying to work out the answer for the temperature T from the equation...

(1012)2=(2.8*1019)(1.04*1019)(T/300)3exp(-1.12/KT)

2. Relevant equations

the equation was derived from ni2=NcNv(T/300)3exp(-Eg/KT)

3. The attempt at a solution

Only simplified the equation to $$\frac{10^{24}}{2.912*10^{38}}$$=$$\left(\frac{T}{300}\right)$$3 . exp$$\left(\frac{-1.12}{KT}\right)$$

can someone advice how to simplify this equation to solve for T?

Tried shifting (T/300)^3 over to the left side and then taking ln on both sides but could not solve the equation as I still ended up having trouble with grouping the T terms due to ln ...
Attached attempted solution.

#### Attached Files:

• ###### Untitled-1.pdf
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2. Sep 25, 2009

### ehild

You can not express T with a closed formula in this case, but you can find its approximate value by trial and error, or with a graphical method.
I can not open your pdf file, so I do not see what you have done. I wonder if you used appropriate value for the Boltzman constant, K. Eg is given certainly in electron volts, so you should use the value K=8.6173*10-5eV/K.
It might be a good method to take the logarithm of both sides, and arrange the equation so that T is on the left side and ln(T) and everythig else on the other. Than you start with T=300 K, insert it into the formula on the right side of the equation, and calculate the new T. Now repeat the procedure with this value. Try!

ehild

3. Sep 29, 2009

### snoothie

Yup got it. Thanks.

Seems like there isn't any easier ways. no?

Did take log on both sides to try making the equation easier to work with, but since trial and error is the way to solve this problem. I found it easier to just plug in no.s into $$\frac{10^{24}}{2.912*10^{38}}$$ = $$\left(\frac{T}{300}\right)^3$$ . exp$$\left(\frac{-1.12}{KT}\right)$$ and just work on T from here...

Last edited: Sep 29, 2009
4. Sep 29, 2009

### ehild

Well, yes, you can try and try... But what about the iteration method I suggested?

If you simplify the numerical terms, take the logarithm, express T, you get something like the following:

$$T=\frac{12997 }{16.194+3\ln{T}}$$

Starting with T=300, the next values are 390, 382, 382. Try, it is really fun :)

ehild

5. Sep 29, 2009

### ehild

I could see your pdf file at last. Have you got problems with ln(T3)? You certainly remember that ln(a*b)=ln(a)+ln(b). T3 is just T*T*T, so what about replacing ln(T3) with 3*ln(T)???

ehild

6. Sep 29, 2009

### snoothie

oh silly me... I was only thinking about addition and subtraction and did not think about shifting T over to the left side and shifting all the terms from left side to the right side of the equation as the denominator of -12997. I kept thinking goodness how to work this with ln|9.27*10^-8|-3ln|T|... and got myself stuck with ln|9.27*10^-8|-3ln|T|=(-1.12/KT)...

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