Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The electron concentration in an $n$ -type semiconductor is the same as hole concentration in a $p$ -type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.

Current in $n$ -type = current in $p$ -type

Current in $p$ -type > current in $n$ -type

Current in $n$ -type > current in $p$ -type

No current will flow in $p$ -type, current will only flow in $n$ -type

Step-by-Step Solution

The current through a semiconductor is $I = neAv_d = neA\mu E$ . The ratio of currents is $\frac{I_n}{I_p} = \frac{n_e e A \mu_e E}{n_h e A \mu_h E} = \frac{\mu_e}{\mu_h}$ . Since $\mu_e > \mu_h$ , therefore $I_n > I_p$ .

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