The Correct Answer is Option (D)

Currently no explanation available

The Correct Answer is Option (A)

Option A is the correct statement about Zener diode. It works as a voltage regulator in reverse bias and behaves like a simple pn junction diode in forward bias. When a Zener diode is reverse-biased, it operates in the breakdown region, where a relatively constant voltage is maintained across the diode, regardless of the current flowing through it. This property makes it useful as a voltage regulator. In forward bias, the voltage applied across the diode is in the same direction as the normal direction of current flow. In this condition, the Zener diode behaves like a simple pn junction diode and allows current to flow in the forward direction.

The Correct Answer is Option (C)

As temperature increases$n_e$ increases, this results in increase in conductance.

$\therefore T$ increases,$n_e$ increases and $R$ decreases.

The Correct Answer is Option (C)

Statement I is correct but in statement II we cannot detect the current through ammeter thus the statement II is incorrect.

The Correct Answer is Option (A)

Pentavalent activities have excess free e^$-$, so e^$-$ density increases but overall semiconductor is neutral.

The Correct Answer is Option (D)

In n-type semiconductor pentavalent impurity is added. Each pentavalent impurity donates a free electron. So the Fermi-level of n-type semiconductor will go upward.

And In p-type semiconductor trivalent impurity is added. Each trivalent impurity creates a hole in the valence band. So the Fermi-level of p-type semiconductor will go downward.

The Correct Answer is Option (C)

For semiconductor,

Conductivity, $\sigma$ = n_e q $\mu$_e + n_h q $\mu$_h

given that semiconductor is n-type. So contribution of holes is ignored.

$\therefore$   n_h q $\mu$_h = 0

$\therefore$   $\sigma$ = n_e q $\mu$_e

Resistivity, $\rho$ = $\frac{1}{\sigma }$

= $\frac{1}{n_{e}q{\mu }_e}$

= $\frac{1}{{10}^{19}\times 1.6\times {10}^{-19}\times 1.6}$

= $\frac{1}{1.6\times 1.6}$

= 0.4 $\mathrm{\Omega }$ m

The Correct Answer is Option (A)

Conductivity of semiconductor,

$\sigma$ = e$\left({\eta }_e{\mu }_e+{\eta }^{\prime }{\mu }_h\right)$

= 1.6 $\times$ 10^$-$19 (5 $\times$ 10¹⁸ $\times$ 2 + 5 $\times$ 10¹⁹ $\times$ 0.01)

= 1.6 $\times$ 1.05

= 1.68

The Correct Answer is Option (A)

The Correct Answer is Option (A)

$Si$ and $Ge$ are semiconductors but $C$ is an insulator. Also, the conductivity of $Si$ and $Ge$ is more than $C$ because the valence electrons of $Si,Ge$ and $C$ lie in third, fourth and second orbit respectively.

The Correct Answer is Option (C)

$\frac{I_e}{I_h}=\frac{n{}_{e}eA{v}_e}{n{}_{h}eA{v}_h}$

$\Rightarrow \frac{7}{4}=\frac{7}{5}\times \frac{v_e}{v_h}$

$\Rightarrow \frac{v_e}{v_h}=\frac{5}{4}$