Question: 13. 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.
(1) current in n-type = current in p-type.
(2) current in p-type > current in n-type.
(3) current in n-type > current in p-type.
(4) No current will flow in p-type, current will only flow in n -type.
Answer: Option (3)
Explanation:
The current density in a semiconductor due to charge carriers is given by
J = nq\mu E, where n is the carrier concentration,
q is the charge, \mu is the mobility
and E is the applied electric field.
In an n-type semiconductor, the majority charge carriers are electrons,
and the current density is J_n = n e \mu_n E.
In a p-type semiconductor, the majority charge carriers are holes,
and the current density is J_p = p e \mu_p E.
Given that the electron concentration in the n-type semiconductor is equal to the hole concentration in the p-type semiconductor, we have n = p.
The mobility of electrons \mu_n is greater than the mobility of holes \mu_p.
Since current is directly proportional to mobility, J_n > J_p.
Therefore, the current in the n-type semiconductor is greater than the current in the p-type semiconductor, and the correct answer is Option (3).