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(D) The total current density $J$ is the sum of the current density due to electrons $(J_e)$ and the current density due to positive ions $(J_i)$.$J =

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$0.1$

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(D) The total current density $J$ is the sum of the current density due to electrons $(J_e)$ and the current density due to positive ions $(J_i)$.$J = J_e + J_i = n e v_e + n e v_i = n e (v_e + v_i)$Given:$n = 5 	imes 10^7\, cm^{-3} = 5 	imes 10^7 	imes 10^6\, m^{-3} = 5 	imes 10^{13}\, m^{-3}$$e = 1.6 	imes 10^{-19}\, C$$v_e = 0.4\, m/s$$J = 4\, \mu A/m^2 = 4 	imes 10^{-6}\, A/m^2$First,calculate the current density due to electrons:$J_e = n e v_e = (5 	imes 10^{13}) 	imes (1.6 	imes 10^{-19}) 	imes 0.4 = 8 	imes 10^{-6} 	imes 0.4 = 3.2 	imes 10^{-6}\, A/m^2$Now,find the current density due to ions:$J_i = J - J_e = 4 	imes 10^{-6} - 3.2 	imes 10^{-6} = 0.8 	imes 10^{-6}\, A/m^2$Using $J_i = n e v_i$,we solve for $v_i$:$v_i = \frac{J_i}{n e} = \frac{0.8 	imes 10^{-6}}{(5 	imes 10^{13}) 	imes (1.6 	imes 10^{-19})} = \frac{0.8 	imes 10^{-6}}{8 	imes 10^{-6}} = 0.1\, m/s$.

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