Question

An electron has drift velocity of $4\times {10}^{–4}\text{m/s}$. If the mobility of electron is $3\times {10}^{–2}{\text{m}}^2{\text{V}}^{–1}{\text{s}}^{–1}$, then applied electric field is -

• A. $4\times {10}^{-6}{\text{Vm}}^{-1}$
• B. $7.5\times {10}^{-6}{\text{Vm}}^{-1}$
• C. $1.33\times {10}^{-2}{\text{Vm}}^{-1}$
• D. $3.33\times {10}^{-9}{\text{Vm}}^{-1}$

Answer 1

The correct option is C $1.33\times {10}^{-2}{\text{Vm}}^{-1}$

Mobility of electrons is equal to drift velocity per unit electric field,

$\mu =\frac{v_d}{E}$

$\therefore E=\frac{v_d}{\mu }=\frac{4\times {10}^{-4}}{3\times {10}^{-2}}=1.33\times {10}^{-2}{\text{Vm}}^{-1}$