I Can plotting corrected values help determine a more precise E_g?

[Kara386]

I have the equation

$qV = E_g + T[kln(\frac{I}{A}) - (3 + \frac{\gamma}{2})kln(T)]$

So if I plot $qV$ against $T$ that'll be a straight line with the y-intercept being $E_g$. But then my lab manual says a more precise value of $E_g$ can be found by plotting the corrected value

$qV_c = qV + (3 + \frac{\gamma}{2})kTln(T)$

So does that mean if I want to plot $qV_c$ against $T$, which I think is what's being asked, then I should plot

$qV_c = E_g + kTln(\frac{I}{A})$

And I'll get my more accurate $E_g$?

 

[BvU]

Hi,

You can't plot $E_g$ to find $E_g$. (which probably is also not how I should understand your last line (*))

Your first

So if I plot $qV$ against $T$ that'll be a straight line

ignores the $- (3 + \frac{\gamma}{2})kln(T)$ in the full $qV$ expression. I suppose this is a small disturbance (?) and $qV_c$ is an attempt to get a more linear relationship.

You really want to plot $qV_c$ against $T$ ((*) which I think is how I should understand your last sentence).

But of course, the proof of the pudding is in the eating. Does it indeed yield a more linear graph ?

PS if you want to avoid things like $kln$ in $\TeX$ use \ln to get $k\ln$

 

[Kara386]

Hi,

You can't plot $E_g$ to find $E_g$. (which probably is also not how I should understand your last line (*))

Your first
ignores the $- (3 + \frac{\gamma}{2})kln(T)$ in the full $qV$ expression. I suppose this is a small disturbance (?) and $qV_c$ is an attempt to get a more linear relationship.

You really want to plot $qV_c$ against $T$ ((*) which I think is how I should understand your last sentence).

But of course, the proof of the pudding is in the eating. Does it indeed yield a more linear graph ?

PS if you want to avoid things like $kln$ in $\TeX$ use \ln to get $k\ln$

Sorry, should have made it clearer. :) As you thought, I meant I'll plot the corrected value of $qV_c$ against $T$ and the intercept of that graph will be my more accurate $E_g$. I think I can consider the $- (3 + \frac{\gamma}{2})kln(T)$ to be constant for $T$ in the range 200 to 400 Kelvin, where all measurements will be taken, but probably the correction tries to deal with inaccuracies caused by that assumption. And thanks for the tip, I'll switch to $\ln$ in future!

 

[pixel]

Plotting your first equation vs. T will not give a straight line as there is a ln(T) term. Plotting

$qV_c = qV + (3 + \frac{\gamma}{2})kTln(T)$

vs. T will give a straight line since T is now appearing linearly on the right hand side of the equation. The intercept of that straight line on the vertical axis will be E_g.

 

[Kara386]

Plotting your first equation vs. T will not give a straight line as there is a ln(T) term. Plotting

$qV_c = qV + (3 + \frac{\gamma}{2})kTln(T)$

vs. T will give a straight line since T is now appearing linearly on the right hand side of the equation. The intercept of that straight line on the vertical axis will be E_g.

If I sub $qV = qV_c - (3 + \frac{\gamma}{2})kT\ln(T)$ into my first equation, that gives

$qV_c = E_g + kT\ln(\frac{I}{A})$

Which is linear in $T$. Are you saying that's what I should plot, or just plot

$qV_c = qV + (3 + \frac{\gamma}{2})kTln(T)$?

 

[BvU]

How about doing it instead of asking ?

 

[ZapperZ]

If I sub $qV = qV_c - (3 + \frac{\gamma}{2})kT\ln(T)$ into my first equation, that gives

$qV_c = E_g + kT\ln(\frac{I}{A})$

Which is linear in $T$. Are you saying that's what I should plot, or just plot

$qV_c = qV + (3 + \frac{\gamma}{2})kTln(T)$?

Please note that only this equation

qV_c = E_g + kT\ln(\frac{I}{A})

will give you a linear equation in T if you plot qV_c versus T. All the others that you wrote have extra "T" hanging around elsewhere in the equation.

Zz.

 

[Kara386]

How about doing it instead of asking ?

I don't actually have any data yet. And I know only one of the equations is linear, therefore must be the right one, so checking which one to plot was unecessary except for my peace of mind, I suppose. Bad habit of mine. Thanks for your help everyone! :)