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In the free-electron model, the drift velocity acquired by electrons in a steady field E (with $\tau$ the relaxation time, e and m the electron charge and mass) has magnitude
A$v_d = \dfrac{eE\tau}{m}$
B$v_d = \dfrac{mE}{e\tau}$
C$v_d = \dfrac{eEm}{\tau}$
D$v_d = \dfrac{e\tau}{mE}$
Answer & Solution
Correct answer: A. $v_d = \dfrac{eE\tau}{m}$
1. An electron accelerates as $a = eE/m$ between collisions.
2. Over the average time $\tau$ between collisions, $v_d = a\tau$.
3. Substituting, $v_d = eE\tau/m$.
4. The other forms misplace $m$, $\tau$, or $E$ and give wrong dimensions.
_Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.5_
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