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Quick Physics Practical Question.

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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:

--- Quote from: AH2 on June 03, 2009, 11:08:42 pm ---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!

--- End quote ---


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...............
 :D

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?

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