Quick Physics Practical Question.

[bottlerockets]

I was going over the old practicals and saw the one from October/November 2003. (specifically: 9702/05/O/N/03) and am stuck on part (f) of question 1.

We plotted ln I/A on the y-axis and V on the x-axis and the equation relating the two given is:

I = Io e^(eV/kT)

That is e raised to the power eV/kT and it's Io multiplied by e.


I'm not a mathematics student, so I have no idea how to solve this equation into y=mx+c and figure out what is equal to m and what is equal to c. I've never done any calculations with ln before.

If anyone could help me with this, I'd be very much obliged. Thank you so much.

[AH2]

Hey bottlerockets, I just saw ur question I think this is the answer:
I=Ioe^ev/kt
u then divide both sides by Io to get:
I/Io=e^ev/kt
u then (Ln) both sides to get:
Ln(I/Io)=ev/kt
In logarithms if ur havin Ln(x/y), then itz the same as Lnx-Lny, so:
LnI-LnIo=ev/kt
To make the equation in y=mx+c form:
LnI=ev/kt+LnIo
y=LnI
m=e/kt
c=LnIo
Hope this helps, sorry if the symbols are not clear!

[bottlerockets]

Hey bottlerockets, I just saw ur question I think this is the answer:
I=Ioe^ev/kt
u then divide both sides by Io to get:
I/Io=e^ev/kt
u then (Ln) both sides to get:
Ln(I/Io)=ev/kt
In logarithms if ur havin Ln(x/y), then itz the same as Lnx-Lny, so:
LnI-LnIo=ev/kt
To make the equation in y=mx+c form:
LnI=ev/kt+LnIo
y=LnI
m=e/kt
c=LnIo
Hope this helps, sorry if the symbols are not clear!

This helped alot!
Thank you so much.
You're a lifesaver, my friend. :]

[Edward10]

Hey guys what is the gradient in the 1st Q ?
I think it is in the range between 5 - 10........
also tell the y intercept...............
 

[AH2]

I think I got mine wrong:
Gradient: 3.2 somethin like that
Intercept: 0
wat did u get for the 1st time (t) using 50cm?
I got 19s is it right?