Question

The magnitude of the drift velocity is very small compared to the speed of random electron motion in a metal. The mean time between collisions can be calculated from the mean free path d, which is the average distance that an electron can travel before colliding with one of the metal nuclei. Using these variables, what is the mean time between collisions τ? Let EF be the energy of the electrons.
Express the mean time between collisions in terms of EF, d, and m.

Answer 1

ik velocity is equal to e*Electric field*mean time between collissions/mass of electron and mean time of collision equals distance (d)/velocity but i dont know how to express mean time betweeen collisions with distance(d) mass(m) and the final electric field(EF)

Answer 2

sorry i cant help in physics

Answer 3

Express the mean time between collisions in terms of EF, d, and m.

Answer 4

This should help you get started

Within materials,
$$
F=eE\text{ force on electron}\\
a={F\over m_e}\text{acceleration of electron }\\
v_d=a\tau\text{ drift velocity, where }\tau \text{ is time between collision and }\\
\text{a is acceleration}\\
v_d={eE\over m_e}\tau\\
$$

Need to go, will check back later

Answer 5

none of these equations have a d in them. I need my final answer to be in terms of distance too

Answer 6

$$
v_d={eF\tau \over m_e}\\
\tau = {m_ev_d\over eF}
$$

Answer 7

From this I believe you can put in terms of d. Will be back in a few hours

Answer 8

is it the square root of md/eF which is the square root of md/e^2E. I got that by plugging in d/t for velocity

Answer 9

or if u do t=square root(2d/a) it could be the square root of (2dm/eE). Not sure which answer is right

Answer 10

I also get something similar -

We can also express drift velocity as
$$
v_d={d\over \tau}\\
$$
So then
$$
F=eE\text{ force on electron}\\
a={F\over m_e}\text{acceleration of electron }\\
v_d={d\over \tau}={eE\over m_e}\tau\\
\tau^2={d~m_e\over e~E}\\
\tau=\sqrt{{d~m_e\over e~E}}=\sqrt{{d~m_e\over F}}
$$

Answer 11

that's incorrect when I type it in for hw I need an expression in terms of just ef d and m

Answer 12

If by \(E_F\) you mean the Fermi Energy, then
$$
v_d={eEd\over m_e}\times \sqrt {m\over 2E_F}={eE\over m_e}\tau\\
\implies \tau = d\sqrt{m\over2E_F}
$$
This represents \(\tau\) in terms of all the parameters you specified.

References:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/fermi.html#c1