Types Semiconductors (P type and N type) Questions in English

(B) In a semiconductor,the charge carriers responsible for the flow of electric current are both free electrons (in the conduction band) and holes (in the valence band).
When an external electric field is applied,these charge carriers undergo drift,resulting in an electric current.
Therefore,the current in a semiconductor is due to the drift of both free electrons and holes.

(C) In a $p-$type semiconductor,the material is doped with trivalent impurity atoms (like Boron,Aluminum,etc.).
These trivalent atoms create vacancies in the valence band of the semiconductor crystal lattice.
These vacancies are known as holes.
Since the number of holes created by the doping process is significantly higher than the number of thermally generated electrons,holes act as the majority charge carriers in $p-$type semiconductors.

(B) $P-$ type semiconductor is formed by doping an intrinsic semiconductor (like Silicon or Germanium) with a trivalent impurity atom.
Gallium $(Ga)$ belongs to Group $13$ of the periodic table and has a valency of $3$.
When Gallium is added to pure Silicon (a Group $14$ element),it creates a deficiency of electrons,known as a hole,which acts as a charge carrier,resulting in a $P-$ type semiconductor.
Arsenic,Antimony,and Phosphorus are pentavalent impurities (Group $15$) and would result in an $N-$ type semiconductor.

(D) To convert an intrinsic semiconductor like germanium $(Ge)$ into a $P-$type semiconductor,we must add an impurity atom that has fewer valence electrons than germanium.
Germanium is a group $14$ element with $4$ valence electrons.
By adding a trivalent impurity (an element from group $13$,such as Boron,Aluminum,or Gallium) which has $3$ valence electrons,we create a hole in the crystal lattice.
This hole acts as a charge carrier,resulting in a $P-$type semiconductor.
Therefore,the valence of the impurity is $3$.

(B) Given:
Electron concentration $(n_e)$ = $8 \times 10^{14} \text{ cm}^{-3}$
Hole concentration $(n_h)$ = $5 \times 10^{12} \text{ cm}^{-3}$
Since $n_e > n_h$,the number of free electrons is significantly greater than the number of holes.
In a semiconductor,if the concentration of electrons is greater than the concentration of holes,it is classified as an $N$-type semiconductor.
Therefore,the correct option is $B$.

(B) In a $P$-type semiconductor,trivalent impurity atoms are doped into an intrinsic semiconductor (like $Si$ or $Ge$).
These trivalent atoms have three valence electrons,which form covalent bonds with three neighboring atoms.
The fourth bond remains incomplete,creating a vacancy known as a 'hole'.
This hole represents the absence of an electron,which acts as a positive charge carrier with a magnitude equal to the electronic charge.
Therefore,the correct option is $(b)$.

(B) To create an $N$-type semiconductor,we need to add a pentavalent impurity (an element with $5$ valence electrons) to an intrinsic semiconductor like germanium $(Ge)$.
When a pentavalent atom (such as phosphorus,arsenic,or antimony) is added,$4$ of its valence electrons form covalent bonds with the neighboring germanium atoms,while the $5$th electron remains free to conduct electricity.
Therefore,the valence of the impurity atom must be $5$.

(A) When a pentavalent impurity like Arsenic $(As)$ is added to Silicon $(Si)$,it introduces free electrons into the crystal lattice.
This process,called doping,creates an $n$-type semiconductor.
The addition of these extra charge carriers significantly increases the electrical conductivity of the material.

(A) To obtain a $P$-type semiconductor,we must dope the intrinsic semiconductor $(Si)$ with a trivalent impurity atom.
$Si$ belongs to group $14$ of the periodic table and has $4$ valence electrons.
Aluminium $(Al)$ belongs to group $13$ and has $3$ valence electrons.
When $Al$ is added to $Si$,it creates a hole (vacancy) in the crystal lattice,which acts as a charge carrier,resulting in a $P$-type semiconductor.
Phosphorous is a pentavalent impurity used for $N$-type semiconductors.

(C) An $N-$ type semiconductor is electrically neutral.
During the doping process,pentavalent impurity atoms (like Phosphorus or Arsenic) are added to the intrinsic semiconductor (like Silicon or Germanium).
Since the impurity atoms themselves are electrically neutral and the host semiconductor crystal is also neutral,the resulting $N-$ type semiconductor remains electrically neutral overall,despite having an excess of free electrons.

(B) Semiconductors,whether $N-$type or $P-$type,are electrically neutral.
During the process of doping,neutral impurity atoms are added to the intrinsic semiconductor crystal.
Since the total number of protons equals the total number of electrons in the added impurity atoms and the host crystal atoms,the net charge of the resulting semiconductor material remains zero.

(C) Germanium $(Ge)$ is a tetravalent semiconductor element belonging to group $14$ of the periodic table.
Phosphorus $(P)$ is a pentavalent impurity atom belonging to group $15$ of the periodic table.
When a pentavalent impurity is added to a tetravalent semiconductor,it provides an extra electron for conduction.
This results in an excess of electrons,creating an $N$-type semiconductor.
Therefore,the correct option is $C$.

(C) In an intrinsic semiconductor,the material is pure and contains no dopants.
Thermal energy at room temperature causes some valence electrons to break their covalent bonds and move to the conduction band,leaving behind a vacancy known as a hole.
Since each electron-hole pair is generated simultaneously,the number of free electrons $({n_e})$ must be equal to the number of holes $({n_h})$.
Therefore,${n_e} = {n_h}$ in an intrinsic semiconductor.

(B) To create a $P$-type semiconductor,a pure semiconductor like silicon (which is tetravalent) must be doped with trivalent impurity atoms.
These trivalent atoms have $3$ valence electrons.
When they replace a silicon atom in the crystal lattice,they create a vacancy or 'hole' because they can only form $3$ covalent bonds with the surrounding silicon atoms.
Among the given options,Boron $(B)$ is a trivalent element (Group $13$ element),while Phosphorus $(P)$ and Antimony $(Sb)$ are pentavalent (Group $15$ elements) and Copper $(Cu)$ is a transition metal.
Therefore,Boron is the correct impurity used to make a $P$-type semiconductor.

(D) In intrinsic semiconductors,both electrons and holes act as charge carriers due to thermal excitation.
In $P-$ type semiconductors,holes are the majority charge carriers.
Therefore,holes are charge carriers in both intrinsic and $P-$ type semiconductors.
Thus,the correct option is $(d)$.

(D) In extrinsic semiconductors,the doping concentration is typically very low compared to the number of host semiconductor atoms.
Typically,for every $10^{6}$ to $10^{8}$ host atoms,one impurity atom is added.
Therefore,the ratio of impurity atoms to pure semiconductor atoms is approximately $10^{-6}$ to $10^{-8}$.
Among the given options,$10^{-7}$ is the most appropriate value representing this ratio.

(B) In a semiconductor,a hole represents the absence of an electron in the valence band. When an electron moves from the valence band to the conduction band or is captured by an impurity atom,it leaves behind a vacancy in the covalent bond structure. This vacancy is referred to as a hole,which acts as a positive charge carrier. Therefore,a hole in a $P-$type semiconductor is a missing electron.

(D) In $P$-type semiconductors,the doping is done with trivalent impurities,which create an abundance of holes in the valence band.
Therefore,holes are the majority charge carriers.
Due to thermal excitation,a small number of electrons are also present in the conduction band,which act as the minority charge carriers.
Thus,the correct order is holes and electrons.

(C) Germanium is a group $14$ element with $4$ valence electrons.
Antimony $(Sb)$ is a group $15$ element,which means it has $5$ valence electrons.
When antimony is added to germanium as a dopant,$4$ of its valence electrons form covalent bonds with the surrounding germanium atoms.
The $5$th electron remains loosely bound and becomes a free charge carrier.
Since the added impurity provides extra electrons,it acts as a donor impurity,resulting in an $N$-type semiconductor.
In an $N$-type semiconductor,the number of free electrons is significantly greater than the number of holes at room temperature.

(D) An extrinsic semiconductor is formed by adding impurities to an intrinsic semiconductor. There are two types of extrinsic semiconductors:
$1$. $P$-type semiconductor: Formed by adding trivalent impurity atoms. In this case,holes are the majority charge carriers,so $N_h \gg N_e$.
$2$. $N$-type semiconductor: Formed by adding pentavalent impurity atoms. In this case,electrons are the majority charge carriers,so $N_e \gg N_h$.
Therefore,the relationship between $N_h$ and $N_e$ depends on the type of impurity added to the semiconductor. Thus,option $(d)$ is correct.

(B) Indium $(In)$ is a group $13$ element,which means it is trivalent (has $3$ valence electrons).
Germanium $(Ge)$ is a group $14$ element,which is tetravalent (has $4$ valence electrons).
When trivalent Indium is added as an impurity to tetravalent Germanium,it creates a deficiency of electrons in the crystal lattice,known as a hole.
Since holes act as positive charge carriers,the resulting semiconductor is a $P-$ type semiconductor.

(C) Antimony $(Sb)$ is a pentavalent element (Group $15$ of the periodic table).
When antimony atoms are added as an impurity to intrinsic germanium (a Group $14$ element),they provide extra electrons to the crystal lattice.
Since the impurity atoms donate electrons to the conduction band,the resulting material is an $N$-type semiconductor.

(A) To create an $N-$type semiconductor,a pentavalent impurity (an element from Group $15$ of the periodic table) must be added to an intrinsic semiconductor like Germanium $(Ge)$.
Arsenic $(As)$ is a pentavalent element,meaning it has $5$ valence electrons.
When Arsenic is doped into Germanium,$4$ of its valence electrons form covalent bonds with the surrounding Germanium atoms,and the $5$th electron becomes a free charge carrier,contributing to the $N-$type conductivity.
Therefore,Arsenic is the correct impurity.

(D) In an $N-$type semiconductor,the majority charge carriers are electrons and the minority charge carriers are holes.
When the semiconductor is heated,thermal energy provides enough energy to break covalent bonds in the crystal lattice.
Each broken bond creates an electron-hole pair.
Therefore,the number of electrons and the number of holes both increase by the same amount due to thermal generation.
Thus,the correct option is $(d)$.

(C) To obtain a $P$-type semiconductor,the intrinsic semiconductor (like Germanium or Silicon) must be doped with a trivalent impurity atom.
Among the given options,Arsenic $(As)$,Antimony $(Sb)$,and Phosphorus $(P)$ are pentavalent elements (Group $15$),which are used to create $N$-type semiconductors.
Indium $(In)$ is a trivalent element (Group $13$),which creates a deficiency of electrons (holes) when added to Germanium,thus forming a $P$-type semiconductor.
Therefore,the correct option is $C$.

(C) $P-$type semiconductor is formed by doping an intrinsic semiconductor (like $Si$ or $Ge$ from Group $14$) with a trivalent impurity (from Group $13$).
$Al$ (Aluminum) and $B$ (Boron) are trivalent elements.
$As$ (Arsenic) and $P$ (Phosphorus) are pentavalent elements,which form $N-$type semiconductors.
Therefore,mixing $Al$ in $Si$ (Option $B$) and $B$ in $Ge$ (Option $C$) results in $P-$type semiconductors.
Thus,the correct combination is $B$ and $C$.

(A) In a $P-$type semiconductor,the doping is done using trivalent impurities,which create an excess of holes.
Since holes are the majority charge carriers,the current in a $P-$type semiconductor is mainly carried by holes.
Therefore,option $A$ is correct.

(D) In an $N$-type semiconductor,pentavalent impurity atoms (like Phosphorus or Arsenic) are added to the intrinsic semiconductor (like Silicon or Germanium).
These impurity atoms provide extra electrons to the conduction band.
Therefore,in $N$-type semiconductors,electrons are the majority charge carriers,while holes are the minority charge carriers.

(A) Phosphorus is a pentavalent impurity (Group $15$ element).
When a semiconductor is doped with a pentavalent impurity,it becomes an $n$-type semiconductor.
In an $n$-type semiconductor,the number of electrons $({n_e})$ is much greater than the number of holes $({n_h})$ because the impurity atoms donate extra electrons to the conduction band.
Therefore,${n_e} \gg {n_h}$.

(C) To obtain an $N-$type semiconductor,pure silicon (a group $14$ element) is doped with a pentavalent impurity (group $15$ element),such as Phosphorus $(P)$.
To obtain a $P-$type semiconductor,pure silicon is doped with a trivalent impurity (group $13$ element),such as Indium $(In)$.
Therefore,the correct combination is Phosphorus for $N-$type and Indium for $P-$type.

(D) To obtain an $N-$ type semiconductor,an intrinsic semiconductor like germanium (a group $14$ element) must be doped with a pentavalent impurity (a group $15$ element).
Pentavalent impurities have $5$ valence electrons.
When added to the germanium crystal lattice,$4$ electrons form covalent bonds with neighboring germanium atoms,and the $5$th electron becomes a free charge carrier (electron).
Both Phosphorus $(P)$ and Arsenic $(As)$ are pentavalent elements.
Therefore,doping germanium with either Phosphorus or Arsenic results in an $N-$ type semiconductor.
Thus,the correct option is $(d)$.

(D) To obtain electrons as majority charge carriers in a semiconductor,we need to create an $n$-type semiconductor.
This is achieved by doping an intrinsic semiconductor (like $Si$ or $Ge$) with pentavalent impurity atoms (such as $P$,$As$,or $Sb$).
These pentavalent atoms have $5$ valence electrons,where $4$ electrons form covalent bonds with the host atoms and the $5$th electron becomes free to conduct electricity,thus making electrons the majority charge carriers.

(C) $P$-type semiconductors are formed by doping an intrinsic semiconductor (like Silicon or Germanium) with trivalent impurity atoms. Among the given options,Boron $(B)$ is a trivalent element (Group $13$),while Arsenic $(As)$,Phosphorus $(P)$,and Bismuth $(Bi)$ are pentavalent elements (Group $15$). Therefore,adding Boron creates holes,resulting in a $P$-type semiconductor.

(B) In an intrinsic semiconductor,thermal energy at room temperature is sufficient to excite electrons from the valence band to the conduction band.
For every electron excited to the conduction band,a corresponding hole is created in the valence band.
Therefore,the number density of electrons $(n_e)$ is equal to the number density of holes $(n_h)$,such that $n_e = n_h = n_i$,where $n_i$ is the intrinsic carrier concentration.

(C) To form an $N$-type semiconductor,a pentavalent impurity (an element from Group $15$) must be added to an intrinsic semiconductor like $Si$ (Silicon).
$As$ (Arsenic) is a pentavalent element,meaning it has $5$ valence electrons.
When $As$ is doped into $Si$,$4$ of its valence electrons form covalent bonds with $Si$ atoms,and the $5$th electron becomes a free charge carrier,contributing to $N$-type conductivity.
$Al$ (Aluminum) and $B$ (Boron) are trivalent impurities (Group $13$) and are used to create $P$-type semiconductors.

(C) The process of intentionally adding small amounts of specific impurities to an extremely pure (intrinsic) semiconductor to modify its electrical properties is known as doping.
By adding these impurities,the concentration of charge carriers (electrons or holes) increases,which significantly enhances the conductivity of the semiconductor.
Therefore,the correct option is $C$.

(B) Germanium $(Ge)$ is a tetravalent semiconductor. Phosphorus $(P)$ and Antimony $(Sb)$ are both pentavalent elements (Group $15$ elements). When a pentavalent impurity is added to a tetravalent semiconductor,it provides an extra electron for conduction. This results in the formation of an $N$-type semiconductor.

(B) Germanium $(Ge)$ is a group $14$ element with $4$ valence electrons.
Gallium $(Ga)$ is a group $13$ element,which means it has $3$ valence electrons.
When a trivalent impurity like Gallium is added to a tetravalent semiconductor like Germanium,it creates a vacancy or a 'hole' in the crystal lattice.
Since the majority charge carriers in this doped semiconductor are holes (positive charge carriers),the material behaves as a $P-$ type semiconductor.

(B) Donor impurities are elements that have $5$ valence electrons,known as pentavalent elements (e.g.,Phosphorus,Arsenic,Antimony).
When these atoms are doped into a semiconductor like Silicon or Germanium,$4$ of their valence electrons form covalent bonds with the host atoms,while the $5$th electron remains free to conduct electricity.
Since these atoms donate an extra electron to the conduction band,they are called donor-type impurities.

(B) In semiconductor physics,a hole is defined as the absence of an electron in the valence band.
Since an electron has a negative charge of $-e$,the absence of an electron creates a vacancy that acts as a positive charge carrier.
The magnitude of the charge on a hole is exactly equal to the magnitude of the charge on an electron,which is $+e$.
This is equivalent to the charge of a proton,which is also $+e$.
Therefore,the charge on a hole is equal to the charge of a proton.

(C) Germanium $(Ge)$ is a tetravalent semiconductor. Phosphorus $(P)$ is a pentavalent impurity (donor atom).
When $Ge$ is doped with $P$,each $P$ atom provides one extra free electron to the crystal lattice.
These free electrons act as negative current carriers.
Therefore,the doped material has more negative current carriers (electrons) compared to holes,making it an $n$-type semiconductor.
Since the crystal remains electrically neutral as a whole,option $(c)$ is the most appropriate description of the carrier concentration.

(A) In a $p$-type semiconductor, the concentration of holes $(n_h)$ is approximately equal to the concentration of acceptor atoms $(N_A)$.
Given: $n_h \approx N_A = 10^{21} /m^3$.
The intrinsic concentration of electron-hole pairs is $n_i = 10^{19} /m^3$.
Using the law of mass action for semiconductors: $n_e \cdot n_h = n_i^2$.
Substituting the values: $n_e \cdot 10^{21} = (10^{19})^2$.
$n_e \cdot 10^{21} = 10^{38}$.
$n_e = 10^{38} / 10^{21} = 10^{17} /m^3$.
Therefore, the concentration of electrons is $10^{17} /m^3$.

(A) In an $N$-type semiconductor,pentavalent impurity atoms are added to a pure semiconductor (like $Si$ or $Ge$).
These impurity atoms provide extra electrons to the conduction band.
The energy levels of these donor electrons are called donor energy levels $(E_d)$.
These donor energy levels are located just below the conduction band $(E_c)$,typically at a distance of about $0.01 \ eV$ for $Ge$ and $0.05 \ eV$ for $Si$.
Therefore,the donor level lies closely below the bottom of the conduction band.

(D) To form a $P-$type semiconductor,an intrinsic semiconductor like germanium $(Ge)$ must be doped with a trivalent impurity atom.
Gallium $(Ga)$,Boron $(B)$,and Aluminium $(Al)$ are all elements of Group $13$ in the periodic table,meaning they have $3$ valence electrons.
When these atoms are added to germanium,they create holes in the crystal lattice,resulting in $P-$type conductivity.
Therefore,all the given options are correct.

(D) An extrinsic semiconductor,whether it is $N$-type or $P$-type,is formed by adding a small amount of impurity to a pure (intrinsic) semiconductor.
Although the impurity atoms provide extra charge carriers (electrons or holes),the impurity atoms themselves are electrically neutral before they are added to the crystal lattice.
When these atoms are incorporated into the lattice,the total number of protons equals the total number of electrons in the entire crystal structure.
Therefore,the net charge of an extrinsic semiconductor remains zero,making it electrically neutral.

(A) In a semiconductor,as the temperature increases,more covalent bonds break,leading to an increase in the number density of charge carriers $(n_e)$. Thus,$n_e$ increases.
However,the drift velocity $(v_d)$ is given by the relation $v_d = \mu E$,where $\mu$ is the mobility. Mobility is inversely proportional to the relaxation time $(\tau)$,which decreases as temperature increases due to more frequent collisions. Therefore,the mobility $\mu$ decreases,causing the drift velocity $v_d$ to decrease.
Thus,$n_e$ increases and $v_d$ decreases.

(B) In an $N$-type semiconductor,the donor impurity energy level lies just below the conduction band $(CB)$.
This allows electrons to be easily excited from the donor level to the conduction band,increasing the conductivity of the semiconductor.
Looking at the provided diagrams,option $B$ correctly illustrates the impurity energy level positioned just below the conduction band $(CB)$.

(B) Mobility $(\mu)$ is defined as the drift velocity per unit electric field, given by $\mu = \frac{e\tau}{m^*}$, where $\tau$ is the relaxation time and $m^*$ is the effective mass. Electrons have a smaller effective mass compared to holes and experience less scattering within the crystal lattice. Consequently, electrons have a higher relaxation time and lower effective mass, leading to higher mobility than holes.

(D) The conductivity $\sigma$ of an $N$-type semiconductor is given by the formula: $\sigma = e n_e \mu_e$,where $e$ is the elementary charge,$n_e$ is the electron concentration (impurity concentration),and $\mu_e$ is the electron mobility.
Given: $\sigma = 6.24 \ mho/cm$,$\mu_e = 3900 \ cm^2/V \cdot s$,and $e = 1.6 \times 10^{-19} \ C$.
Rearranging the formula to solve for $n_e$:
$n_e = \frac{\sigma}{e \mu_e}$
Substituting the values:
$n_e = \frac{6.24}{1.6 \times 10^{-19} \times 3900}$
$n_e = \frac{6.24}{6240 \times 10^{-19}}$
$n_e = \frac{6.24}{6.24 \times 10^{-16}}$
$n_e = 10^{16} \ cm^{-3}$.

(B) The correct answer is $B$ $(N-type)$.
When a pentavalent impurity (donor impurity) is added to an intrinsic semiconductor like Silicon or Germanium,it provides extra free electrons to the crystal lattice.
These electrons act as majority charge carriers.
Since the charge carriers are negatively charged electrons,the resulting semiconductor is known as an $N-type$ semiconductor.