Question

In a semiconductor, it is found that $\frac{3}{5}th$ of the total current is carried by electrons and the remaining $\frac{2}{5}th$ by the holes. If at this temperature, the drift speed of electrons is $2.5$ times that of holes, the ratio of the number densities of electrons and holes is

• A. $\frac{3}{5}$
• B. $\frac{5}{8}$
• C. $\frac{25}{8}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{3}{5}$

Given: $I_e=\frac{3}{5}x$
$I_h=\frac{2}{5}x$
$x$ = total current
$I_h,Ie$ = current due to holes and electrons respectively.
$Ve=2.5V_h$
Now, we know $I=nVe$
n = number density
v = drift speed
e = charge on electron = charge on hole
So, $n=\frac{I}{Ve}$
$\frac{n_e}{n_h}=\frac{Ie\times V_h}{Ve\times I_h}=\frac{3x\times V_h\times 5}{5\times 2.5V_h\times 2x}$
$=\frac{3}{5}\times \frac{5}{2}\times \frac{10}{25}$
$\frac{n_e}{n_h}=\frac{3}{5}$.