post office box experiment viva questions

Determine Unknown Resistance by a Post Office Box

Experiment: To determine the unknown resistance by a Post Office Box

Theory: By inserting appropriate resistance in the post office box current passing through the galvanometer can be made zero. If the potentials between points B and D are equal i.e., there is no potential difference between the two terminals of a galvanometer, then this phenomenon happens. In this situation, no deflection is found in the galvanometer and the bridge remains in the equilibrium condition. Now according to Wheatstone bridge balanced condition,

S = R x Q/P

If P = 10 ohm and Q = 10 ohm

So, S = (10 x R) / 10 = R ohm

Again by putting P = 100 ohms and Q = 10 ohms, if the experiment is performed then it will be seen,

100/10 = R/S

So, S = 10R/100 = 1/10 R ohm.

Similarly taking P = 1000 ohm and Q = 10 ohms, if the experiment is performed, then we get 1000/10 = R/S

So, S = 1/100 x R ohm … … … (1)

By this instrument, the value of unknown resistance can be determined up to the second digit after the decimal. Here R = resistance of the 3rd arm.

Description of the instrument

A Post Office Box can also be used to measure an unknown resistance. It is a Wheatstone bridge with three arms; while the fourth arm(s) is the unknown resistance. Post office box (in short, PO box) is the special form of Wheatstone bridge. It is called a post office box because it was originally used by the postal department for the measurement of the resistance of the telegraph wires and cables.

Normally, with the help of this instrument, we can measure both low and high resistances. It is composed of some fixed value resistance coils. Coils are arranged in succession and form three arms of the Wheatstone bridge. The experimental unknown resistance S is the fourth arm of the bridge. Like the resistance box, here also coils are in a box. On the top platform of the box, two ends of each coil are connected by solid brass blocks. Brass blocks an arranged in adjacent three rows. Between two successive blocks, there is an arrow-shaped gap. If the plug of the gap is withdrawn fixed resistance is included in the circuit.

In Figure, the schematic diagram of the post office box is shown. In this instrument, three sections of coil resistances of varying magnitudes are connected by plugs and are marked as AC. CD and AE which constitute respectively the three arms P, Q, and R of the Wheatstone bridge. The arms P and Q are called ratio arms. Each of if contains three resistors of 10, 100 and 1000 Ω values connected in series. Third arm R normally consists of resistances from 1 Ω to 5000 Ω connected in series. When the plug of any coil is taken out, its resistance will be included in the circuit. An unknown resistance S is connected between points D and E which forms the fourth arm of the Wheatstone bridge. A battery B is connected between points A and D through a tapping key K 1 and a galvanometer is connected between points C and E through another tapping key K 2 .

The lower portion of the key K 1 and the point A is connected internally. Similarly, point C and the lower portion of the key K 2 is connected internally. So, try the key current in the main point and by the key K 2 current in the galvanometer can be made on and off.

Procedure :

(1) By rubbing the ends of connecting wires and contact points thoroughly connection is made tightly according to the figure.

(ii) From each arm of P and Q, 10-ohm resistance plugs are taken out. Keeping the resistance of the third arm R zero, first the key of the battery circuit K 1 is pressed and then the key K 2 of the galvanometer circuit is pressed and the deflection of the galvanometer is observed. Now withdrawing the ∞ (infinity) plug the deflection of the galvanometer is observed again. In this case, if the deflection of the galvanometer is opposite to the first deflection, then it is considered that the circuit connection is alright.

(iii) Starting from the higher values of resistance of the third arm deflections of the galvanometer are noticed by reducing the value of resistance progressively. When the galvanometer gives zero deflection for P = Q = 10 ohms, the unknown resistance S = R will be equal to the third arm resistance. If the deflection is not made zero, then due to the two successive resistances the deflection will be opposite. In this situation, the value of the unknown resistance S is between these values.

(iv) Now in arm P instead of 10 ohms, 100-ohm resistances are inserted. As before the resistance of the second arm, Q is kept at 10 ohms. So Q/P ratio will be 1/10, hence in balanced condition R = 10 S. Now as before resistance plugs from the third arm R are withdrawn so that galvanometer deflection becomes zero. In this situation, unknown resistance S = R/10.

(v) Similarly, 1000 ohm resistance is inserted in arm P and in Q arm 10-ohm resistance is inserted and the experiment is performed as before.

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post office box experiment in physics | practical

To verify laws of resistances in series by post office box.

To verify laws of resistances in series, a post office box is used to measure unknown resistance. It works on the principle of Wheatstone bridge . P, Q, R and S be the four resistances of the Wheatstone bridge as shown in figure 1. Here, P, Q and R are the known resistances and S is the unknown resistance. A galvanometer is connected between B and D. When the galvanometer current is zero (null condition ), the relation between P, Q, R and S is \frac{P}{Q} = \frac{R}{S} S= \frac{Q}{P}R Where, S is the equivalent resistance of two resistances connected in series. P and Q are the resistances in the ratio arm and R is the resistance of the rheostat arm. Or, S= r_{1}+r_{2}

Procedure for post office box experiment or practical in physics (Need not to write in your laboratory notebook)

• Two known resistances are connected in series.
• All circuit elements are connected as shown in Figure 2.
• To check the correctness of the circuit, first 10 \Omega resistance is put in the two ratio arms i.e. P and Q. A zero resistance is put in the third arm. Then the battery key is pressed first and after that the galvanometer key is pressed. Notice the direction of galvanometer scale deflection (say left)
• Now make the resistance of the third arm infinity or maximum. Check the direction of galvanometer scale deflection after pressing the two keys. If the deflection of the galvanometer is opposite (say right) to the previous one, then the connection is correct.
• \frac{P}{Q} =\frac{R}{S} \frac{10}{10} =\frac{R}{S} R=S So, the unknown resistance will be in between R and R+1.
• Now, set the resistances in the ratio arm as 100:10. So, \frac{P}{Q} =\frac{R}{S} \frac{100}{10} =\frac{R}{S} R=10S So, R will be 10 times of S. Find two resistances in the third arm R 1 and (R 1 +1) to make the galvanometer deflection very low and opposite.
• Therefore, the unknown resistance will be in between \frac{R_{1}}{10} and \frac{R_{1}+1}{10} .
• Now, set the ratio arm as 1000:10. \frac{P}{Q} =\frac{R}{S} \frac{1000}{10} =\frac{R}{S} R=100S So, R will be 100 times of S.
• Find two resistances in the third arm R 2 and (R 2 +1) to make the galvanometer deflection very low and opposite
• Therefore, the unknown resistance will be in between \frac{R_{2}}{100} and \frac{R_{2}+1}{100} .

Table for finding equivalent resistance in series connection

Precautions or Discussions

• The battery key should be pressed first and then the galvanometer key.
• All plugs should be filled tightly.
• Low currents should be passed to the P.O box.

Viva questions on post office box experiment in physics | practical (TO VERIFY LAWS OF RESISTANCES IN SERIES BY POST OFFICE BOX)

1. why is the instrument called p.o (post office ) box .

Ans : This instrument is used to measure the unknown resistances of electric cable and telegraphic wires. It is used by the postal department to find resistance of communication wires. This is why the name of this instrument is post office (P.O) box.

2. In which principle does the post office box works?

Ans: It works on the principle of Wheatstone bridge.

3. State the principle of Wheatstone bridge.

Ans: The condition at which the galvanometer deflection is zero is called null condition. At null condition, the relation between the four resistances P, Q, R and S is \frac{P}{Q} =\frac{R}{S}

4. How do you know the connection is correct.

Ans : See procedure no. (3)

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• Current Electricity
• Measurement Of Unknown Resistance Using A Post Office Box
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Current Electricity of Class 12

Measurement of unknown resistance using a post office box.

Now, the resistances in P and Q are made 100 Ω and 1000 Ω, respectively, and the process is repeated.

Equation (1) is used to compute S.  The ratio P/Q is progressively made 1:10, and then 1:100. Thus, the resistance S can be accurately measured.

The major sources of error are the connecting wires, unclear resistance plugs, change in resistance due to Joule heating, and the insensitivity of the Wheatstone bridge.

These errors may be removed by using thick connecting wires, clean plugs, keeping the circuit on for very brief periods (to avoid Joule heating) and calculating the sensitivity.

In order that the sensitivity is maximum, the resistance in the arm P is kept close to the value of the resistance S.

Illustration14. In an experiment with a post−office box, the ratio arms are 1000 : 10.  If the value of the third resistance is 999 Ω,. find the unknown resistance.  

Solution: The ratio arms are 1000 : 10 

∴ P/Q = 1000/10 = 100

Third resistance R = 999 Ω

Let x be the unknown resistance. 

We know that,

∴ X = Q/P x R = 1/100 x 999 = 9.99 Ω

Illustration15.When two resistances X and Y are put in the left hand and right hand gaps in a meter bridge,  the null point is at 60 cm. If X is shunted by a resistance equal to half of itself then find the shift in the null point.

⇒I = 33.3 ∴ Shift  = 60 – 33.3 = 26.7 cm

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Measurement of an unknown resistance using a Post Office Box

A Post Office Box can also be used to measure an unknown resistance. It is a Wheatstone Bridge with three arms P, Q and R; while the fourth arm(s) is the unknown resistance. P and Q are known as the ratio arms while R is known as the rheostat arm.

 At balance, the unknown resistance

S = (P/Q) R                                                           …… (1)

The ratio arms are first adjusted so that they carry 100 W each. The resistance in the rheostat arm is now adjusted so that the galvanometer deflection is in one direction. If R = R0 (ohm) and R = R0 + 1 (ohm) are the resistance in rheostat arm, for which the deflection in galvanometer is in opposite direction, then it implies that the unknown resistance ‘s’ lies between R0 & (R0 + 1) ohm.

Alt txt: resistance-using-post-office-box

Now, the resistance in P and Q are made 100 W and 1000 W, respectively, and the process is repeated. Equation (1) is used to compute S. The ratio P/Q is progressively made 1:10, and then 1:100. Thus, the resistance S can be accurately measured.  

 The major sources of error are the connecting wires, unclear resistance plugs, change in resistance due to Joule heating, and the insensitivity of the Wheatstone bridge

These errors may be removed by using thick connecting wires, clean plugs, keeping the circuit on for very brief periods (to avoid Joule heating) and calculating the sensitivity.

In order that the sensitivity is maximum, the resistance in the arm P is kept close to the value of the resistance S.

Illustration:

 Resistance of a conductor is 1.72 W at a temperature of 20°C. Find the resistance at 0oC and 100°C. Given the coefficient of resistivity is a = 0.00393/oC.

R = R0 (1 + aDT)

R = 1.72 W (1 + 0.00393) × (0°C – 20°C) = 1.58 W

At T = 100°C

R = 1.72 W [1 + 0.00393) × (100°C – 20°C)] = 2.26 W

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To Determine the Temperature Coefficient of Resistance for platinum using Carey Foster’s bridge and Platinum Resistance Thermometer (PRT)

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5 thoughts on “To Determine the Temperature Coefficient of Resistance for platinum using Carey Foster’s bridge and Platinum Resistance Thermometer (PRT)”

🙏 Namste sir, I can’t under stand how to calculate the resistance R1,R2,R3 using (l1-l2). and the temperature co efficient of platinum resistance thermometer is 0.00385/°C

🙏 Namste sir, 4.9×10^(−7) Ω/m is the resistivity of constantan wire

Have you seen my video on this expeirment on my youtube channel all lab experiments

yes, but i can’t understand the formula of the platinum resistance thermometer Rt=r+ρ(L1-L2)

sir how to find error analysis

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Post Office Box Physics || Current Electricity Class 12, JEE & NEET

• At balance, the unknown resistance S = (P/Q) R                                                           …… (1)
• It is based on principle of Wheatstone bridge.
• Unknown resistance is $\mathrm{S}=\frac{\mathrm{Q}}{\mathrm{P}} \mathrm{R}$ and specific resistance is $\rho=\frac{\pi r^{2} S}{L},$ where $r$ is radius and $L$ is length of wire.

• It is used to find unknown resistance, specific resistance of a wire, internal resistance of cell, resistance of galvanometer etc.
• A typical post office box is in a wooden box with a hinged lid and a metal or Bakelite panel showing circuit connections. Coils of wire are wound non-inductively, mounted in the body of the box, and have a negligible temperature coefficient. Pairs of ratio arms are each 10, 100, 1000 ohms. A resistance arm contains a number of coils from 1 to 5000 ohms with a plug for infinite resistance.
• Types and Effects of Electric Current
• Ohm’s Law and Resistance
• Combination of Resistances
• EMF and Internal Resistances of a Cell
• Cells Connected in Series, parallel and Mixed
• Kirchhoff’s Circuit Law
• Electric Currents in Conductors
• Wheatstone Bridge
• Post office Box
• Wheatstone Meter Bridge
• Moving Coil galvanometer
• Ammeter and Voltmeter
• Potentiometer Working Principle

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• class 12 practical to verify the laws of combination series of resistances using a metre viva questions

Physics Practical Class 12 - To verify the laws of combination (series) of resistances using a metre bridge. Viva Questions with Answers

1) What is a metre bridge?

A slide wire bridge, also termed a metre bridge, is an instrument that works on the principal Wheatstone bridge. To find unknown resistance of a conductor, a metre bridge is used.

2) In a series combination of resistance, how do you find the equivalent resistance?

Since the same current passes through each resistor in series combination, the total resistance R T can be calculated using the below equation:

R T = R 1 + R 2 + R 3 + …….R n

3) What is Wheatstone bridge?

A Wheatstone bridge is a particular type of electrical circuit that is used in measuring the unknown electrical resistance of the circuit by balancing the two legs of the bridge circuit, where the unknown component includes one of its legs.

4) What is the use of a metre bridge?

To measure the resistance precisely for a resistor, a metre bridge is used.

5) Which principle is followed by the metre bridge for its working?

The principle of the Wheatstone bridge is followed by a metre bridge for its efficient working.

6) Which material wire is used in a metre bridge?

The materials such as nichrome, constantan or manganin are used in making the wire of a metre bridge because these materials have a high value of resistance, and the coefficient of the temperature of their resistances is low.

7) What is the case when the metre bridge is in a more sensitive condition?

The metre bridge circuit will be very sensitive when all four resistors have the same resistance values.

8) What is the balanced condition of a Wheatstone bridge?

When no current flows through the galvanometer, the Wheatstone bridge is said to be in a balanced condition. By adjusting the known resistance and variable resistance, this condition can be achieved.

9) What is the working principle of a Wheatstone bridge?

It is the principle of null deflection which is responsible for the working of a Wheatstone bridge, i.e. no current flows through the circuit, and the ratio of their resistances are equal.

10) What are the restrictions of Wheatstone bridge?

The resistance of the leads and contacts becomes important for low resistance measurement, but the Wheatstone bridge shows errors while measuring them.

• Practical Procedure
• Measurement of Resistance
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Experiment-1: Name of the Experiment: Determination of the value of an unknown resistance by means of a post office box

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The net amount of charge flowing through a cross section in unit time is defined as current. Mathematical Equation: If rate of flow of charge is independent of time, then the electric current is said to be steady then I = Q/t If rate of flow of charge varies with time, the current at any time i.e. instantaneous current is given by I = dQ/dt. If n be the number of conduction electrons crossing cross –section in time t then I = ne/t. SI unit of current is ampere. One ampere is defined as If one coulomb of charge flows across any of its cross-section in one second. Note: Electric current has direction as well as magnitude but it is not a vector quantity. This is because currents do not add like vectors. Solved Numerical Q) An electric device sends out 78 coulombs of charge through a conductor in 6 seconds. Find current flow Solution Given Q = 78C, time of flow t = 6s The current I = Q/t = 78/6 = 13A Q) What is the quantity of electricity required to provide a current o 10A for one hour Solution Given current I = 10A, time of flow t = 1 hour = 3600s The quantity of electricity = amount of charge flowing Q = It Q = (10)(3600) = 36000C Q) The current through a wore varies with time as I = IO + αt , where Io = 10A and α = 4 A/s. Find the charge that flows across a cross-section of the wire in first 10 seconds Solution: Current I = dq/dt = IO + αt dq = (IO + αt)dt Integrating on both sides ∫ í µí±‘í µí±ž = ∫ (í µí°¼ 0 + í µí»¼í µí±¡) í µí±¡=10 í µí±¡=0 í µí±‘í µí±¡ í µí±ž = [í µí°¼ 0 í µí±¡ + í µí»¼í µí±¡ 2 2 ] í µí±¡=0 í µí±¡=10 q= 10IO +50α substituting IO = 10 and α = 4 q= 10(10) + 50(4) = 300C

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Planck’s Constant Experiment Excellent 15 Viva Questions

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Last updated on Sunday, October 15th, 2023

Planck’s Constant Experiment Viva Questions:

Table of Contents

Planck’s Constant using LED

Determination of planck’s constant by using light-emitting diodes (leds)..

APPARATUS: Planck’s constant kit, wires, graph paper, and 3-4 LEDs.

FORMULA USED: Planck’s constant is h = eV /ν, where e is an electronic charge, V is the voltage reading in a voltmeter, and v is the frequency of a particular LED color.

THEORY:    The energy of a photon is given by the equation: E = h ν                           (1)

Where E is the energy of a photon, ν is its frequency, and ( h) is Planck’s constant.

CASE 1: In the case of the photoelectric effect:

• An electron is emitted from the metal only if the energy of the incident photon is greater than the work function of the metal.
• These electrons can be attracted by the anode in a circuit and as a result of a current, you can observe with respect to the incident radiations (color).
• If you need to measure the voltage difference, you can use resistance (the choice is open) in the circuit.
• This voltage will be corresponding to the particular incident radiation energy.
• If you need to know any other parameters, you can use this information (the current, the voltage, the color of incident radiation means the wavelength or frequency, so you can find the E=hv of the incident photon).

READ ALSO: Semiconductor Diode Laser Viva

Case 2.  in the case of leds, the opposite of the above-mentioned working is true..

• Here, in LED, If an electron of sufficient electrical energy (eV) is passed across a material then a photon emits.
• But remember, the meaning of passing the electron across a material here is a diode.
• That has two types of semiconductors (n- and p-type) and a p-n junction. Near the p-n junction, there is a specific region known as the depletion region.
• These electrons start from the n-region and after crossing the barrier (depletion region) reach the p-region where they recombine and as a result, a photon emits.
• But to understand it you have to understand the energy band concept, this explanation is given in the semiconductor laser topic.

Planck’s Constant Experiment Supporting Concepts to LED

Please note that all materials don’t show the photoelectric effect and emission of radiation like in LED. For LED we use Ga As the material that shows the optical properties when electron-hole recombination takes place. If you got the flavor of the second case then it will be clear that you need forward biasing for this purpose. When you provide a sufficient voltage to the electron to cross the barrier only then it recombines with holes. And only then you can see the photons means that light. So initially you can not see the light when you provide potential to the LED. But when you reach the threshold value of voltage where the electron is able to cross the junction then only you see the light. This value of potential you know is known as stopping potential. Now the point is clear, also the emitted photon energy (hv) will be the same as the electrical energy of the electron  (eV). Because of this reason you use; eV   =   h ν                                        h   = eV / ν This equation we will use to determine Planck’s constant.

PLANCK’S EXPT. PROCEDURE:

• Make the connection in the kit.
• Take the current measurement of each LED by varying the voltage as given in the table.
• Plot the curve on the graph paper between the frequency of color on the X-axis and electrical energy on Y-axis for all LEDs.
• The slope of the curve will give a measured value of Planck’s constant.
• Compare the measured value with the standard value and check the percentage error.

OBSERVATION S:

As mentioned in the procedure plot a graph between the last two-column of the above table that is the frequency of the particular LED and energy (eV). Take the slope of this graph and this will be your measured value of Planck’s constant. Now compare it with the standard value (6.62607004 × 10 -34  m 2 kg / s) and explain the percentage error. Check yourself that what can be the reasons for this percentage error.

What you can analyze in Planck’s Constant Experiment?

From your observation, you can also analyze the stopping potential of all the LEDs. How? Just see you have taken readings for each LED, either that one when the LED starts to glow or also after it with some intervals of potential. So you have a set of reading with potential and current for each LED. When you will plot it for every LED you will observe that every LED starts with a specific value of the potential. This value of the potential is known as the stopping potential and this way, you can analyze the stopping potential for each color LED. But to determine Planck’s constant you will need a graph that points explained above.

How to Observe Planck’s Constant from this Experiment?

• Take current and voltage readings for each LED (Red, Green, and Blue)
• Plot a graph on the graph paper using proper scaling of the variables (Voltage and Current), as shown in the above figure.
• Now find the knee voltage for each one, and make one more graph between stopping potential and frequency as shown in 2nd part of the above picture.
•  Take the slope from here and use it in the formula as you can see right side of the above figure.
• Calculate the percentage error by using the formula and check the reasons for it, if more than 20%.

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Points of Care when You make the Graphs:

• When you are plotting the graph between any two physical variables of the experiment, first make a scale along the X-axis and Y-axis. For example, 10 small boxes are equal to 1 volt along the Y-axis or 1 mA along the X-axis.
• Then mark the points where it lies
• After it joins all the points. Now at this point keep remembering, it is not necessary that all your points will lie in a straight line. You have to draw a line that passes over to or near each point. Like, I have drawn the line above, which is passing near to that point which was away from it.
• Now you can take any two points on the line except for them which were marked earlier. Find the slope and put it in the required formula. This will be your observed value.
• I hope it will be enough to get my point, for any doubt you can ask through the comment box.

VIVA QUESTIONS Planck’s Constant Experiment

• How Planck’s constant is determined? “Planck’s constant is determined by using the light-emitting diode. The LED works on the principle of semiconductor diode, to flow the current from n to p-type semiconductor a minimum potential is required. When electron reaches the p-type semiconductor they fuse there and emits a photon. So in this way, we can compare the potential energy eV 0 with the energy of the photons h and nu.

Q1. How it is different from Si/Ge diode?

Ans . LED is made of Ga As Gallium Arsenide semiconductor material which shows the optical properties when electron-hole recombination takes place.  While Si/Ge-made diode or semiconductor shows thermal properties, they start to heat up when current flows. On the other hand LED glow.

2Q. How LED works?

2Answer:   When forward bias gives to the light emitting diode (LED), immediately LED doesn’t glow and takes some time. The minimum potential at which LED starts to glow is known as the stopping potential. The light from LED is the result of electron and hole recombination in the depletion region.

3Q. Why Minimum potential is required to glow the LED?

3Ans. There is a potential barrier for the charge carrier to cross the junction, to overcome it they required this amount of potential energy. And then after with small change in potential, they cross the junction, and current flows through the LED.

4Q. In the photoelectric effect, a suitable frequency of photon falls on an electron in an atom and ejects the electron. In LED when electron-hole recombination takes place a photon emits. How do you see these two phenomena?

4Ans: Both phenomena are different, in the case of the “photoelectric effect” to emit the electron, from the surface of a material, a minimum energy of threshold frequency is required. While on the other hand, in a light-emitting diode (LED) photon emits when electron-hole recombination takes place above the threshold value of the voltage; known as stopping potential.

5Q. Why do you put two different energies eV and hv equal, what is the condition that they satisfy in the LED?

Answer: From the question, no. 3 you understood the stopping potential, and a small potential above it shows the deflection in current, simultaneously glowing in the LED. The potential energy eV is responsible to recombine the electron-hole recombination, by which a photon of the energy hv emits. Because of this reason, one can put eV = hv

6Q.Which material do we use in the LED?

Answer: Gallium Arsenide which is of a semiconductor nature.

7Q. How are photons emitted from the LED and from which section of the LED?

Answer: When electron-hole recombine photons emit, these emit from the depletion region.

8Q. How do you explain the working of LED by using the energy band diagram in forward biasing?

Answer: You have to follow a detailed lecture for this purpose. Click Here Working of LED.

7Q. What happens when you provide the forward bias to the LED in terms of the conduction band & valence band in the depletion region?

Answer: Conduction band and valence band drift in the depletion region, for more info watch this video .

8Q. Why does not LED starts to glow immediately when you provide the forwarding bias to that?

Answer: Because of the potential barrier at the junction for the charge carriers.

9Q. Explain the concept of stopping potential in semiconductor diode by V-I Characteristics.

Answer: In the above figure, it is pointed out for the red, green, and blue LEDs.

10Q. Why does Blue color LED stopping potential greater than the Red color LED?

Answer: from the relation eV 0 =hν

further  ν = c/λ so  eV 0 =hν

eV 0 =hc/λ in this relation e, h, and c are constants, so you can see V 0    ∝1/λ

for small wavelength, stopping potential is higher than the larger wavelength. As you can see in Table. 1 the blue color wavelength is 475 nm while for the red it is 650 nm.

11Q. Can we achieve the population inversion process in LED too? if yes what is the condition for that? if not then why?

Answer: Highly doped semiconductor material is required, the first condition, so not possible in LED. and the second reason is a threshold current beyond that we achieve population inversion in the semiconductor LASER.

12 Q. What symbol do we use for the Light Emitting Diode?

Answer: Similar to the normal diode but with arrows, which indicate the emission of light

—-|>-

13Q. What information do we get from Planck’s Constant, and how one can say that radiation is in a discrete form of energy?

Answer: Energy is in the discrete form that signifies by the photon energy;  hv

Sources/Information Required to Explain/Understand the Experiment :

1. Working of a p-n Junction diode 2. Depletion region Concept/Idea along with potential barrier for Si/Ge 3. Semiconductor Material name that shows the optical property 4. Mainly energy band diagram of the p-n junction diode and how electron transit from n to p side in depletion region in case of forward biasing. 5. The basic idea of electrical and radiation energy so can understand why they keep equal two different energy. 

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