The mobility of free electrons (charge = e, mass = m and relaxation time = τ .) in a metal is proportional to

Question

The mobility of free electrons (charge = e, mass = m and relaxation time = τ .) in a metal is proportional to (A)  (e/m)τ  (B)  (m/e)τ  (C)  (e/mτ )  (D)  (m/eτ )

Tardigrade Admin · Accepted Answer

Drift velocity per unit electric field is called mobility of electron, ie, μ=(vd/E)=(E/ρ n e E) Now, ρ=(m/n e2 τ) ∴ s μ=(n e2 τ/m n e)=(e τ/m)

Q. The mobility of free electrons (charge = $e$ , mass = $m$ and relaxation time = $τ$ .) in a metal is proportional to

$\frac{e}{m} τ$

$\frac{m}{e} τ$

$\frac{e}{m τ}$

$\frac{m}{e τ}$

Drift velocity per unit electric field is called mobility of electron,
ie, $μ = \frac{v _{d}}{E} = \frac{E}{ρ n e E}$
Now, $ρ = \frac{m}{n e ^{2} τ}$
$∴ s μ = \frac{n e ^{2} τ}{mn e} = \frac{e τ}{m}$

Q. The mobility of free electrons (charge = e, mass = m and relaxation time = τ .) in a metal is proportional to

me​τ

em​τ

mτe​

eτm​