To solve this problem, we need to compare the currents in an $n$-type and a $p$-type semiconductor.$\\$Given that the electron concentration in the $n$-type is the same as the hole concentration in the $p$-type.$\\$Let's denote:$\\$• $n_e$ as the electron concentration in the $n$-type semiconductor.$\\$• $p_h$ as the hole concentration in the $p$-type semiconductor.$\\$• $\mu_e$ as the electron mobility.$\\$• $\mu_h$ as the hole mobility.$\\$• $E$ as the applied electric field.$\\$The current density $J$ in a semiconductor is given by:$\\ J = q \cdot n \cdot \mu \cdot E \\$For the $n$-type semiconductor:$\\ J_n = q \cdot n_e \cdot \mu_e \cdot E \\$For the $p$-type semiconductor:$\\ J_p = q \cdot p_h \cdot \mu_h \cdot E \\$Since $n_e = p_h,$ we have:$\\ J_n = q \cdot n_e \cdot \mu_e \cdot E \\ J_p = q \cdot n_e \cdot \mu_h \cdot E \\$The ratio of the currents is:$\\ \frac{J_n}{J_p} = \frac{q \cdot n_e \cdot \mu_e \cdot E}{q \cdot n_e \cdot \mu_h \cdot E} \\ = \frac{\mu_e}{\mu_h} \\$Typically, the mobility of electrons $(\mu_e)$ is greater than the mobility of holes $(\mu_h). \\$Therefore, $\frac{\mu_e}{\mu_h} > 1. \\$This implies that $J_n > J_p. \\$Thus, the current in the $n$-type semiconductor is greater than the current in the $p$-type semiconductor. \\ This corresponds to Option 3.