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(B) The mobility $\mu$ of an electron is defined as the ratio of drift velocity $V_d$ to the electric field $E$,given by the formula: $\mu = \frac{V_d

Vedclass · Accepted Answer

$1.6 	imes 10^{-3} \,m^2/V \cdot s$

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(B) The mobility $\mu$ of an electron is defined as the ratio of drift velocity $V_d$ to the electric field $E$,given by the formula: $\mu = \frac{V_d}{E} = \frac{e 	au}{m}$.Given values are:Charge of electron $e = 1.6 	imes 10^{-19} \,C$Mass of electron $m = 9.1 	imes 10^{-31} \,kg$Average collision time (relaxation time) $	au = 9.1 	imes 10^{-15} \,s$.Substituting these values into the formula:$\mu = \frac{(1.6 	imes 10^{-19} \,C) 	imes (9.1 	imes 10^{-15} \,s)}{9.1 	imes 10^{-31} \,kg}$$\mu = \frac{1.6 	imes 10^{-19} 	imes 9.1 	imes 10^{-15}}{9.1 	imes 10^{-31}}$$\mu = 1.6 	imes 10^{-19 - 15 + 31} \,m^2/V \cdot s$$\mu = 1.6 	imes 10^{-3} \,m^2/V \cdot s$.

Find the mobility of an electron in a wire if its average collision time is $9.1 \times 10^{-15} \,s$. (Charge of electron $= 1.6 \times 10^{-19} \,C$ and mass of electron $= 9.1 \times 10^{-31} \,kg$)

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