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Resistivity

Capture.PNG

I did R = pL/A, where p is the resistivity, L = 20 and A = 4 x 20 x d

But the answer is to do p x 20 x d/4 x d^2

Could anyone explain why they have done that please?

Also:
Show that the number density of free electrons within the block is about 2 × 10^28 m^–3. Assume that each phosphorus atom contributes one free electron.

n = 1/d^3, but surely thats only finding the number density in one atom?
(edited 5 years ago)
Original post by Fross877
Capture.PNG

I did R = pL/A, where p is the resistivity, L = 20 and A = 4 x 20 x d

But the answer is to do p x 20 x d/4 x d^2

Could anyone explain why they have done that please?

Also:
Show that the number density of free electrons within the block is about 2 × 10^28 m^–3. Assume that each phosphorus atom contributes one free electron.

n = 1/d^3, but surely thats only finding the number density in one atom?


A is the cross sectional area... the area of a slice orthogonal to the direction of current flow.

I did R=ρL/A and got 224Ω which seems to be what the question is after

A=4 d2 ~ 5.77E-19 m2
L=20d ~ 7.6E-9 m

---
number density is the number of free electrons per m3 so it'd be the same for one atom as for a rectangular block of 80 atoms as shown in the picture.

---
as usual the mark sheme isn't a model answer or an example of the exam board's favourite style.
Original post by Joinedup
A is the cross sectional area... the area of a slice orthogonal to the direction of current flow.

I did R=ρL/A and got 224Ω which seems to be what the question is after

A=4 d2 ~ 5.77E-19 m2
L=20d ~ 7.6E-9 m

---
number density is the number of free electrons per m3 so it'd be the same for one atom as for a rectangular block of 80 atoms as shown in the picture.

---
as usual the mark sheme isn't a model answer or an example of the exam board's favourite style.


Original post by Fross877


I did R = pL/A, where p is the resistivity, L = 20 and A = 4 x 20 x d

But the answer is to do p x 20 x d/4 x d^2

Could anyone explain why they have done that please?

Also:
Show that the number density of free electrons within the block is about 2 × 10^28 m^–3. Assume that each phosphorus atom contributes one free electron.

n = 1/d^3, but surely thats only finding the number density in one atom?


Grrrrrr. :angry: (only kidding) You got there first! lol :smile::smile::smile:

OP: The mark scheme simply used the rectangular face surface area at the 'P' end as a reasonable approximation.

L = 20d

A = d x d x 4 = 4d2

Then

R=ρLA=ρ20d4d2R = \rho\frac{L}{A} = \rho\frac{20d}{4d^2}
(edited 5 years ago)

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