The contribution in the total current flowing | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The total current $I$ in a semiconductor is the sum of electron current $I_e$ and hole current $I_h$. Given $I_e = (3/4)I$ and $I_h = (1/4)I$.Ther

Vedclass · Accepted Answer

$6 : 5$

Vedclass · Answer

(A) The total current $I$ in a semiconductor is the sum of electron current $I_e$ and hole current $I_h$. Given $I_e = (3/4)I$ and $I_h = (1/4)I$.Therefore,the ratio of currents is $I_e / I_h = (3/4) / (1/4) = 3$.We know that current $I = nqvA$,where $n$ is the concentration,$q$ is the charge,$v$ is the drift velocity,and $A$ is the cross-sectional area.Thus,$I_e / I_h = (n_e q v_e A) / (n_h q v_h A) = (n_e / n_h) 	imes (v_e / v_h)$.Given $v_e = (5/2)v_h$,so $v_e / v_h = 5/2$.Substituting the values: $3 = (n_e / n_h) 	imes (5/2)$.Solving for the ratio of concentrations: $n_e / n_h = 3 	imes (2/5) = 6/5$.Thus,the ratio is $6 : 5$.

The contribution in the total current flowing through a semiconductor due to electrons and holes are $3/4$ and $1/4$ respectively. If the drift velocity of electrons is $5/2$ times that of holes at this temperature,then the ratio of concentration of electrons and holes is

Similar Questions

During cell division,sometimes there is a failure of separation of sister chromatids or homologous chromosomes. This event is called

Fraunhofer lines are produced by the absorption of light in

........ of a reaction cannot be determined experimentally.

The value of '$a$' for which the three given planes
$P_1: (a+1)x - y - z = a$
$P_2: x - (a+1)y - z = a$
$P_3: x - y - (a+1)z = a$
have no point in common,is

If the polar of a point on the circle $x^2+y^2=p^2$ with respect to the circle $x^2+y^2=q^2$ touches the circle $x^2+y^2=r^2$,then $p, q, r$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module