Light Emitting Diode (L.E.D), Photodiode, Solar Cell Questions in English

(N/A) The band gap energy of elemental semiconductors, such as silicon $(Si)$ or germanium $(Ge)$, is relatively small. For silicon, the band gap is approximately $1.1 \text{ eV}$, and for germanium, it is about $0.67 \text{ eV}$.
Visible light corresponds to photon energies ranging from approximately $1.8 \text{ eV}$ to $3.1 \text{ eV}$.
Since the band gap of elemental semiconductors is too small, the energy released during electron-hole recombination is insufficient to produce photons in the visible spectrum. Instead, they emit radiation in the infrared region.
Therefore, compound semiconductors with larger band gaps are required to produce visible light $LEDs$.

(C) The band gap energy $E_g$ of a semiconductor is related to the maximum wavelength $\lambda$ of the photon it can detect by the formula: $E_g = \frac{hc}{\lambda}$.
Substituting the given values:
$h = 6.63 \times 10^{-34}\, J \cdot s$
$c = 3 \times 10^{8}\, m/s$
$\lambda = 400 \times 10^{-9}\, m$
$E_g = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^{8})}{400 \times 10^{-9}}\, J$
$E_g = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}}\, J = 4.9725 \times 10^{-19}\, J$.
To convert this energy into electron-volts $(eV)$, divide by the charge of an electron $e = 1.6 \times 10^{-19}\, C$:
$E_g (in\; eV) = \frac{4.9725 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.1\; eV$.

(B) multimeter in resistance mode applies a small $DC$ voltage across the component to measure current.
$(1)$ For a capacitor, the multimeter will show a transient deflection as it charges, regardless of probe polarity. Thus, the statement that it shows $NO$ deflection is incorrect.
$(2)$ An $LED$ is a diode. In forward bias, it conducts (showing deflection), and in reverse bias, it blocks current (no deflection). The option suggests the opposite behavior, making it the observation that will $NOT$ occur.
$(3)$ For a metal wire, the resistance is nearly $0 \ \Omega$, so the multimeter will show $NO$ deflection (or a reading of $0$).
$(4)$ $A$ resistor is an ohmic component; its resistance is independent of the direction of current, so the deflection remains the same when probes are reversed.

(A) photodiode is operated in reverse bias mode. When light of energy greater than the bandgap energy falls on the photodiode,electron-hole pairs are generated in the depletion region.
As the reverse biasing voltage is increased,the electric field in the depletion region increases,which helps in sweeping the generated charge carriers across the junction more effectively,thereby increasing the photocurrent.
However,once all the generated charge carriers are collected,further increase in the reverse bias voltage does not increase the photocurrent significantly,and it reaches a saturation value. This is shown in the $I-V$ characteristic curve of a photodiode.

(D) The energy gap $E_g$ is given as $1.9\, eV$.
To convert this into Joules,we multiply by $1.6 \times 10^{-19} \, J/eV$:
$E = 1.9 \times 1.6 \times 10^{-19} \, J = 3.04 \times 10^{-19} \, J$.
The wavelength $\lambda$ is given by the formula $\lambda = \frac{hc}{E}$.
Substituting the values: $\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3.04 \times 10^{-19}}$.
$\lambda = \frac{19.89 \times 10^{-26}}{3.04 \times 10^{-19}} \approx 6.54 \times 10^{-7} \, m$.
$\lambda = 654 \, nm$.
$A$ wavelength of $654 \, nm$ corresponds to the red region of the visible spectrum.

(C) The energy of the incident photon must be at least equal to the band gap energy $(E_g)$ to generate hole-electron pairs.
The threshold wavelength $\lambda_0$ is given as $621 \, nm$.
The band gap energy is calculated using the formula:
$E_g = \frac{hc}{\lambda_0}$
Using the approximation $hc \approx 1242 \, eV \cdot nm$:
$E_g = \frac{1242 \, eV \cdot nm}{621 \, nm} = 2 \, eV$.
Therefore,the band gap is $2 \, eV$.

(D) photodiode is a $p-n$ junction diode fabricated with a transparent window to allow light to fall on the diode. When the photodiode is illuminated with light (photons with energy $h
u > E_g$),electron-hole pairs are generated due to the absorption of photons. These charge carriers are minority carriers in the depletion region. Under reverse bias,the electric field sweeps these minority carriers across the junction,creating a current. Because the reverse saturation current is highly sensitive to the generation of minority carriers,a small fractional change in the incident light intensity produces a significant,easily detectable change in the reverse bias current.

(A) The energy of a photon emitted by an $LED$ is given by the formula $E = \frac{hc}{\lambda}$.
Given,wavelength $\lambda = 4000 \, \mathring{A} = 400 \, \text{nm}$.
Using the relation $E \approx \frac{1240}{\lambda (\text{in nm})} \, \text{eV}$:
$E = \frac{1240}{400} \, \text{eV} = 3.1 \, \text{eV}$.
Rounding off to the nearest integer,we get $3 \, \text{eV}$.

(A) For a solar cell,the material $A$ must be able to absorb a significant portion of the solar spectrum. The band gap energy of $A$ should be around $1.5 \, eV$ to efficiently convert solar radiation into electrical energy (emf).
Material $B$ is used as a protective coating. To ensure that solar radiation passes through this coating without being absorbed,the band gap energy $E_{B}$ of material $B$ must be significantly larger than the energy of the incident photons. Therefore,a large band gap like $E_{B} = 5 \, eV$ is ideal for material $B$ to remain transparent to the solar spectrum.
Thus,the optimum choice is $E_{A} = 1.5 \, eV$ and $E_{B} = 5 \, eV$.

(C) In a solar cell,when light with energy $hf > E_g$ falls on the $p-n$ junction,electron-hole pairs are generated. These charge carriers are separated by the junction electric field,with electrons moving to the $n$-side and holes moving to the $p$-side.
If no load is connected,these charges accumulate on the $n$ and $p$ sides,creating a photo-voltage.
When an external load is connected,a photo-current $I_L$ flows through the circuit. In this state,the device acts as a power source,delivering power to the external load. The $V-I$ characteristic curve for this operation lies in the fourth quadrant,where the voltage $V$ is positive and the current $I$ is negative (indicating current flowing out of the device). Therefore,the diode generates power in the fourth quadrant.

(B) The correct option is $B$.
Solar cells are devices that convert solar energy into electrical energy. The efficiency of a solar cell depends on the band gap of the semiconductor material used.
$GaAs$ (Gallium Arsenide) has a band gap of approximately $1.43 \,eV$ to $1.5 \,eV$. This value is very close to the peak intensity of the solar radiation spectrum,which makes it an ideal material for absorbing solar energy efficiently.
While $Ge$ (Germanium) and other materials are semiconductors,$GaAs$ is specifically preferred for high-efficiency solar cells due to its optimal band gap alignment with the solar spectrum.

(D) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
For a photodiode to detect light,the energy of the incident photon must be greater than or equal to the band gap energy $(E_g)$ of the semiconductor.
Given $E_g = 3.1 \,eV$.
Using the relation $E \approx \frac{12400}{\lambda (\text{in } \mathring{A})} \,eV$,we set the condition $\frac{12400}{\lambda} > 3.1$.
Solving for $\lambda$,we get $\lambda < \frac{12400}{3.1} \,\mathring{A} = 4000 \,\mathring{A}$.
Since both $4000 \,\mathring{A}$ and $3900 \,\mathring{A}$ satisfy the condition $\lambda \leq 4000 \,\mathring{A}$,both wavelengths can be detected.

(C) In a $p-n$ junction solar cell,when monochromatic light falls on the junction,electron-hole pairs are generated due to the absorption of photons.
These charge carriers are separated by the electric field of the depletion region,creating a potential difference known as the photo-electromotive force (photo-$EMF$).
The number of electron-hole pairs generated is directly proportional to the number of incident photons per unit area per unit time,which is defined as the intensity of the light.
Therefore,the photo-$EMF$ produced is directly proportional to the intensity of the light falling on the cell.

(B) Assertion $A$ is true: Photodiodes are designed to operate in reverse bias. In reverse bias,the reverse saturation current is very small and is highly sensitive to the incident light intensity. This makes it easier to detect small changes in light intensity.
Reason $R$ is true: For a $p-n$ junction diode,the forward bias current is significantly higher than the reverse bias current because the barrier height is reduced in forward bias,allowing majority carriers to flow easily.
However,Reason $R$ is $NOT$ the correct explanation for Assertion $A$. The reason why photodiodes are operated in reverse bias is because the fractional change in the reverse current due to incident light is much larger and easier to measure compared to the forward current,not simply because forward current is larger than reverse current.

(C) Photodiodes are specifically designed to operate under reverse bias conditions to detect light intensity, as the reverse saturation current is highly sensitive to incident light. Therefore, Assertion $A$ is false.
For a $p-n$ junction diode, the forward bias current increases exponentially with voltage once the threshold voltage $V_0$ is reached, whereas the reverse bias current remains very small until the breakdown voltage $V_z$ is reached. Thus, for the range $|V_z| > |V| \geq V_0$, the forward bias current is significantly larger than the reverse bias current. Therefore, Reason $R$ is true.

(C) light emitting diode $(LED)$ is a heavily doped $p-n$ junction diode.
It emits light only when it is forward biased,as the recombination of electrons and holes releases energy in the form of photons.
It does not emit light when reverse biased.
The energy of the emitted light $(E = h
u)$ is approximately equal to or slightly less than the energy band gap $(E_g)$ of the semiconductor material used.
Therefore,statement $C$ is incorrect.

(A) The energy of the incident photon corresponds to the band gap energy $E_g$ of the photodiode.
$E_g = \frac{hc}{\lambda}$
Given $\lambda = 660 \, nm = 660 \times 10^{-9} \, m$, $h = 6.6 \times 10^{-34} \, Js$, and $c = 3 \times 10^8 \, m/s$.
$E_g = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}} \, J$
To convert this energy into $eV$, divide by the elementary charge $e = 1.6 \times 10^{-19} \, C$:
$E_g (in \, eV) = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9} \times 1.6 \times 10^{-19}} \, eV$
$E_g = \frac{19.8 \times 10^{-26}}{1056 \times 10^{-28}} \, eV = \frac{19.8 \times 100}{1056} \, eV = \frac{1980}{1056} \, eV$
Simplifying the fraction: $\frac{1980}{1056} = \frac{15}{8} \, eV$.
Comparing this with $\left(\frac{x}{8}\right) eV$, we get $x = 15$.

(C) The energy of the emitted photon is equal to the band gap energy of the semiconductor, $E_g = 1.42 \,eV$.
The relationship between the wavelength $\lambda$ and the energy $E$ is given by the formula: $\lambda = \frac{hc}{E}$.
Using the approximation $hc \approx 1240 \,eV \cdot nm$, we get:
$\lambda = \frac{1240 \,eV \cdot nm}{1.42 \,eV} \approx 873.24 \,nm$.
Rounding to the nearest given option, we get $\lambda \approx 875 \,nm$.

(D) Statement $A$: $A$ solar cell converts light energy into electrical energy. In the $I-V$ characteristic curve of a solar cell,the voltage is positive and the current is negative (as it delivers power to an external load). This corresponds to the $IV$ quadrant of the $I-V$ graph. Thus,statement $A$ is correct.
Statement $B$: In a reverse-biased $pn$ junction diode,the current is extremely small (in the order of $\mu A$) and is caused by the drift of minority charge carriers across the junction. The majority charge carriers are pushed away from the junction,preventing them from contributing to the current. Thus,statement $B$ is incorrect.

(D) Statement $A$ is incorrect because the junction area of a solar cell is made large to collect more light,whereas a photodiode has a small area for fast response.
Statement $B$ is correct because a solar cell operates in the photovoltaic mode without any external bias.
Statement $C$ is incorrect because an $\text{LED}$ is made of a heavily doped $p-n$ junction to facilitate efficient carrier injection.
Statement $D$ is incorrect because the light intensity of an $\text{LED}$ increases with forward current only up to a certain limit,after which it saturates.
Statement $E$ is correct because an $\text{LED}$ emits light only when it is forward-biased,which causes the recombination of electrons and holes at the junction.
Therefore,statements $B$ and $E$ are correct.

(C) photodiode is operated in reverse bias. When light of sufficient energy is incident on the photodiode,electron-hole pairs are generated in the depletion region.
As the reverse biasing voltage increases,the electric field in the depletion region increases,which helps in sweeping the generated charge carriers across the junction more effectively.
Initially,the photoelectric current increases with the increase in reverse bias voltage.
However,once the voltage is sufficient to collect all the generated charge carriers,the current reaches a saturation value and does not increase further with a further increase in the biasing voltage. This is shown in the characteristic curves for different light intensities $I_1$ and $I_2$.

(A) photodiode is a special type of $P-N$ junction diode that is operated under reverse bias conditions.
When light of energy $h\nu > E_g$ is incident on the junction,electron-hole pairs are generated.
In reverse bias,the electric field at the junction is strong,which sweeps the generated charge carriers across the junction,creating a photocurrent.
The magnitude of this photocurrent is directly proportional to the intensity of the incident light.
Even though the reverse saturation current is very small,it is highly sensitive to the intensity of incident light,making it an ideal device for light detection.
Therefore,both the Assertion and the Reason are true,and the Reason correctly explains the Assertion.

(C) The given characteristic curve shows the $I-V$ relationship in the fourth quadrant,where the device generates power.
Specifically,it shows the 'open circuit voltage' on the $V$-axis and the 'short circuit current' on the $I$-axis.
This behavior is characteristic of a solar cell,which converts light energy into electrical energy and operates in the fourth quadrant of the $I-V$ characteristics curve.
Therefore,the correct option is $C$.

(B) $1$. Element $A$: $A$ capacitor acts as an open circuit for $DC$ (steady state) because its reactance $X_C = 1/(\omega C)$ is infinite at $f = 0$. However,it allows $AC$ to pass through it due to its finite reactance at non-zero frequencies.
$2$. Element $B$: $A$ diode (specifically a $p-n$ junction diode) allows current to flow only in one direction (forward bias) for both $AC$ and $DC$ signals.
$3$. Element $C$: $A$ Light Emitting Diode $(\text{LED})$ is a $p-n$ junction diode that emits light when forward-biased. The intensity of light increases with current (voltage) up to a certain point,after which it may decrease or saturate due to thermal effects or device limitations.
Therefore,$A$ is a capacitor,$B$ is a diode,and $C$ is an $\text{LED}$.

(A) An $\text{LED}$ (Light Emitting Diode) operates under forward bias conditions and emits light when energized.
According to the energy band theory,the energy of the emitted photon is given by $E = h\nu = \frac{hc}{\lambda} \approx E_g$,where $E_g$ is the band gap energy of the semiconductor material.
Since the frequency $\nu$ (or energy) of the emitted light is directly proportional to the band gap energy $E_g$,and the threshold voltage (potential barrier) $V_k$ is also directly related to the band gap energy $(V_k \approx E_g/e)$,it follows that light with a higher frequency requires a higher threshold voltage.
Comparing the frequencies of the colors: $\nu_{\text{blue}} > \nu_{\text{green}} > \nu_{\text{yellow}} > \nu_{\text{red}}$.
Therefore,the threshold voltage required for emission follows the order: $V_{\text{blue}} > V_{\text{green}} > V_{\text{yellow}} > V_{\text{red}}$.
This corresponds to the $I-V$ characteristic curves where the blue light curve is shifted furthest to the right along the voltage axis.

(B) The energy of the incident photon must be greater than or equal to the band gap energy of the semiconductor for detection to occur.
$E_{ph} \geq E_g$
Since $E_{ph} = \frac{hc}{\lambda}$,we have $\frac{hc}{\lambda} \geq E_g$,which implies $\lambda \leq \frac{hc}{E_g}$.
Using the relation $\lambda (\text{in } \mathring{A}) \approx \frac{12400}{E_g (\text{in } eV)}$,we substitute $E_g = 1.4 \ eV$:
$\lambda \leq \frac{12400}{1.4} \ \mathring{A} \approx 8857 \ \mathring{A}$.
Among the given options,the signal with a wavelength of $8000 \ \mathring{A}$ is less than $8857 \ \mathring{A}$,so it can be detected.

(A) In an $LED$,the intensity of emitted light depends on the number of charge carriers crossing the junction. As the forward bias voltage increases from zero,the potential barrier decreases,allowing more electrons and holes to cross the junction and recombine. This recombination releases energy in the form of photons,thus increasing the intensity of the emitted light. Therefore,the intensity increases with the applied forward bias.

(B) An $LED$ (Light Emitting Diode) is a heavily doped $p-n$ junction diode that emits light when biased in the forward direction.
Its $I-V$ characteristics are similar to those of a standard $p-n$ junction diode.
In the forward bias region,the current increases exponentially with the applied voltage after the threshold (knee) voltage is reached.
In the reverse bias region,the current is negligible.
Graph $(a)$ represents a standard $p-n$ junction diode characteristic.
Graph $(b)$ shows a forward bias characteristic starting from the origin,which is the standard representation for an $LED$ in the first quadrant.
Therefore,graph $(b)$ is the correct representation.

(D) photodiode is a special purpose $P-N$ junction diode fabricated with a transparent window to allow light to fall on the diode.
When the photodiode is illuminated with light,electron-hole pairs are generated due to the absorption of photons.
This results in an increase in the reverse saturation current.
The magnitude of the reverse current is directly proportional to the intensity of the incident light.
Looking at the provided $I-V$ characteristics graph,the current increases as we move from $I_1$ to $I_4$ in the reverse bias region.
Since the reverse current is highest for the curve labeled $I_4$,it corresponds to the maximum illumination intensity.
Therefore,the correct option is $D$.

(C) solar cell is a $P-N$ junction device that converts light energy into electrical energy. Option $C$ states that it converts electrical energy into light,which is incorrect. Therefore,it is the mismatch.

(D) photodiode is operated under reverse bias conditions. In circuit $(B)$,the $p$-side is connected to the negative terminal and the $n$-side to the positive terminal,which is reverse bias. Thus,$(B)$ is correct for a photodiode.
An $LED$ (Light Emitting Diode) is operated under forward bias conditions. In circuit $(D)$,the $p$-side is connected to the positive terminal and the $n$-side to the negative terminal,which is forward bias. Thus,$(D)$ is correct for an $LED$.
Therefore,the correct circuit diagrams for normal operation are $(B)$ and $(D)$.

(D) The graph shows current $I$ in $mA$ on the $y$-axis and voltage $V$ in $Volt$ on the $x$-axis in the first quadrant,indicating forward bias characteristics.
Among the given options,a Light Emitting Diode $(LED)$ operates in forward bias,where the current increases exponentially with voltage after the threshold voltage is reached,leading to light emission due to the recombination of charge carriers.
Therefore,the graph represents the forward bias characteristics of an $LED$.

(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
For a semiconductor to detect a signal,the energy of the incident photon must be greater than or equal to the band gap energy $(E_g = 2.5 eV)$.
The threshold wavelength $\lambda$ is calculated as:
$\lambda = \frac{hc}{E_g} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2.5 \times 1.6 \times 10^{-19}} m$
$\lambda = \frac{19.8 \times 10^{-26}}{4.0 \times 10^{-19}} m = 4.95 \times 10^{-7} m = 4950 Å$.
Using the shortcut formula: $\lambda(Å) = \frac{12400}{E(eV)} = \frac{12400}{2.5} = 4960 Å$.
Since the photodiode detects signals with energy $E \ge 2.5 eV$,the wavelength must be $\lambda \le 4960 Å$. Among the given options,$5000 Å$ is the closest threshold,but strictly speaking,the signal must be less than the threshold. Given the standard nature of this problem,$5000 Å$ is the intended answer.

(C) The $I-V$ characteristics of a device operating in the fourth quadrant (where voltage is positive and current is negative,or vice versa depending on convention) represent a device that generates power rather than consuming it.
In a $photocell$,the $I-V$ characteristics are typically shown in the fourth quadrant because it acts as a source of electrical energy when illuminated.
Therefore,the correct option is $C$.

(B) An $LED$ (Light Emitting Diode) is a $p-n$ junction diode that emits light when it is forward-biased.
Therefore,its $V-I$ characteristics are similar to those of a standard $p-n$ junction diode operating in the forward bias region.
In the forward bias region,the current increases exponentially with the applied voltage after the threshold voltage (knee voltage) is reached.
Graph $(b)$ represents this exponential increase in current with voltage in the first quadrant,which is the characteristic behavior of an $LED$ in forward bias.
Graph $(a)$ shows both forward and reverse bias,which is for a general diode.
Graph $(c)$ represents the characteristics of a solar cell.
Graph $(d)$ represents the characteristics of a photodiode.

(D) solar cell is a device that converts solar energy into electrical energy. The solar radiation spectrum has a maximum intensity around $1.5 \ eV$. To efficiently absorb solar radiation,the semiconductor material used in a solar cell must have a band gap that matches this energy range. Materials with a band gap between $1.0 \ eV$ and $1.8 \ eV$ are ideal for this purpose,as they can absorb a significant portion of the solar spectrum. Silicon,with a band gap of approximately $1.1 \ eV$,is the most commonly used material for solar cells.

(B) The wavelength of light emitted by an $LED$ depends on the band gap energy $(E_g)$ of the semiconductor material used.
Aluminium gallium arsenide $(AlGaAs)$ is a semiconductor material with a band gap energy that corresponds to the infra-red region of the electromagnetic spectrum.
Therefore,LEDs manufactured using $AlGaAs$ emit infra-red radiation.

(B) photodiode is a $p-n$ junction diode operated in reverse bias.
In reverse bias,the current is primarily due to the drift of minority charge carriers across the junction.
When light (photons) with energy greater than the bandgap energy falls on the photodiode,it generates additional electron-hole pairs.
These photogenerated minority carriers are swept across the junction by the electric field,thereby increasing the reverse current.
Therefore,the reverse current in a photodiode is directly dependent on the concentration of minority carriers generated by incident light.

(A) The device used for detecting light intensity is a $Photodiode$.
$A$ $Photodiode$ is a special type of $PN$ junction diode that generates current when exposed to light.
It is specifically designed to operate in reverse biased mode.
When light falls on the junction,it creates electron-hole pairs,and the reverse current increases with the intensity of the incident light.
Therefore,it acts as a photodetector or photosensor.

(C) The correct answer is $C$.
$1$. LEDs are highly energy-efficient compared to conventional light sources.
$2$. LEDs have a very long operational lifespan when manufactured with high quality.
$3$. The brightness of light emitted by an $LED$ can be easily controlled by varying the forward current passing through it. Therefore,the statement that it cannot be controlled is incorrect.
$4$. The light emitted by LEDs is monochromatic and does not fade over time.

(D) light emitting diode $(LED)$ is a heavily doped $p-n$ junction diode which emits spontaneous radiation when forward biased.
In the forward biased condition,the potential barrier is reduced,allowing electrons from the $n$-region and holes from the $p$-region to cross the junction.
These charge carriers recombine at the junction,releasing energy in the form of photons (light).
Therefore,an $LED$ is always operated in the forward biased condition to emit light.

(B) The working principle of these devices is as follows:
$(1)$ Photovoltaic cells convert solar energy directly into electrical energy.
$(2)$ Photo-thermal devices absorb solar radiation to produce heat energy.
$(3)$ $A$ Photodiode is a light-sensitive device that generates current when exposed to light,but it is typically used as a detector rather than a primary solar energy harvester.
$(4)$ An $LED$ (Light Emitting Diode) converts electrical energy into light energy,which is the reverse of a solar cell.
Therefore,devices that primarily work on solar energy are the Photovoltaic cell and Photo-thermal devices.

(C) The emission wavelength of an $LED$ depends on the band gap of the semiconductor material used. Zinc selenide $(ZnSe)$ is a wide band gap semiconductor. When an $LED$ is manufactured using zinc selenide,it emits light in the blue region of the visible spectrum.

(D) photodiode is a special type of $p-n$ junction diode that is designed to operate under reverse bias conditions. It is sensitive to light. The symbol for a photodiode consists of a standard $p-n$ junction diode symbol with two arrows pointing towards the diode,indicating that light is incident on it. Since the provided image shows a Zener diode symbol (which has '$Z$' shaped ends on the cathode),it does not represent a photodiode. However,based on standard textbook representations,a photodiode is characterized by incoming light rays.

(D) photodiode is a special type of $p-n$ junction diode that is fabricated with a transparent window to allow light to fall on the diode.
It is operated under reverse bias conditions.
When light with energy $h
u$ greater than the energy gap $E_g$ of the semiconductor is incident on the diode,electron-hole pairs are generated.
Due to the electric field at the junction,these charge carriers are separated before they can recombine,resulting in a current flow in the external circuit.
The magnitude of this photo-current depends on the intensity of the incident light,not significantly on the reverse bias voltage once a certain threshold is reached.
Therefore,the correct statement is that it is always operated in reverse bias.

(C) light emitting diode $(LED)$ is a heavily doped $p-n$ junction diode.
When the diode is forward-biased,electrons from the $n$-region and holes from the $p$-region are injected into the junction region.
In the junction region,these charge carriers undergo recombination.
During the process of recombination,the energy released is in the form of photons.
If the semiconductor material has a suitable band gap,the energy of the emitted photons corresponds to the visible light spectrum.
Therefore,the light emission is due to the recombination of holes and electrons.

(A) photodiode is a semiconductor $p-n$ junction device that is specifically designed to operate under reverse bias conditions. When light of energy greater than the bandgap energy falls on the junction,it generates electron-hole pairs. The reverse bias electric field sweeps these charge carriers across the junction,creating a photocurrent that is proportional to the intensity of the incident light.

(B) light emitting diode $(LED)$ is a heavily doped $p-n$ junction diode which emits light when it is forward biased.
In the schematic symbol of an $LED$,the arrows point away from the diode,indicating the emission of light.
Option $A$ represents a photodiode (arrows pointing towards the diode).
Option $B$ represents an $LED$ (arrows pointing away from the diode).
Option $C$ represents a standard $p-n$ junction diode.
Option $D$ represents a Zener diode.
Therefore,the correct schematic symbol for an $LED$ is shown in option $B$.

(A) When a $p-n$ junction diode is forward biased,the potential barrier is reduced,allowing electrons from the $n$-region and holes from the $p$-region to cross the junction.
At the junction,these charge carriers recombine,and the energy released during this process is emitted in the form of photons.
In diodes made of specific semiconductor materials like gallium arsenide or indium phosphide,this energy corresponds to the visible light spectrum.
Such a device is known as a Light Emitting Diode $(LED)$.

(B) An $LED$ (Light Emitting Diode) is a heavily doped $p-n$ junction diode which emits spontaneous radiation when forward biased. The semiconductor materials used for $LEDs$ must have a band gap energy corresponding to the visible spectrum. Gallium arsenide phosphide $(GaAsP)$ or Gallium phosphide $(GaP)$ are commonly used materials. Among the given options,Gallium arsenide $(GaAs)$ is a standard semiconductor used in $LED$ technology to produce light,although its specific band gap often emits in the infrared,it is the fundamental material class used for these devices.