Q2. Energy density formula for laser using Einstein's coefficient #photonics #physics

Hinzugefügt: 2026-05-26

[00:00:03]Welcome back to my channel. So in this session we're going to be discussing about a derivation on energy density using Einstein coefficient. So you know that in a previous session we have discussed about what do you mean by interaction of light with the matter. So whenever interaction is happening three interaction will be happening. One is induced absorption, another one is spontaneous emission and the third one is stimulated emission. If these three happens in the same stream then the production of laser will be happening. So what is the energy of that particular laser will be derived by using the following expression. So first we'll be finding the rate of induced absorption, rate of spontaneous emission, rate of stimulated emission.

[00:00:52]So let us consider the energy density equation by considering two energy level where E1 is called as a ground state and E2 is called as the exited state. The number of atom which is present inside the energy level E1 is taken as N_sub_1 and the number of atoms which are present in the exited state that is E2 state is taken as N_sub_2. This is the number of atoms at E1 at E2. So what is induced absorption? Normally if I provide a energy so we will be taking it as a E mu. So E mu is called as energy density formula. So the energy density this is the energy which we need to find a derivation by using the phenomena what is happening. So once the energy e mu is provided in any mode of thermal or photon or it might be electric. So the atoms will be going from ground state to exited state. So it will be cons consuming the energy e mu which is directly proportional and it will be having the atoms which will be moving from ground state to exited state is n_sub_1. So if I write this phenomena in this form that rate of induced absorption means how much atoms has been gone per unit cross-sectional area or per unit second will be defined as it will be always proportional to E mu that is energy density and which is the atom which is going the atom will be going N1 atoms will be going for the exited state so the N1 atom so directly proportional if I want to remove this proportionality symbol so I will be adding a constant.

[00:02:46]The constant will be given as B12.

[00:02:50]So what is the indication of 12? 1 2 will be indicating the direction. The atoms are moving from ground state 1 to 2. That is the indication of the direction. So E mu n1 take it as equation number one. So what do you mean by b12? This is a Einstein's coefficient of induced absorption.

[00:03:22]So that is the constant which is given by the Einstein for this particular stream. Similarly you know that after 10 ^ of -8 seconds the whatever the atoms which has been gone for the exited state they should come back for the ground state by liberating the energy here there is no absorption of energy will be taken place so it is directly proportional no energy involved so that's why emu will be not used and the atoms which is coming for the ground state is n2 atoms so here directly n2 atom will be involved if I want to remove the proportionality symbol. So I will be introducing a constant that is EA 21. So 21 indicates the atoms are moving from exalted state to the ground state. So since the emission and the absorption what is happening here is totally different. That's why A21 will be taken as a different constant.

[00:04:17]N_sub_2 will be taken as equation number two. So what is A21?

[00:04:22]It indicates Einstein coefficient of coefficient of spontaneous emission.

[00:04:40]Similarly, there is a third process which will be happening for the atoms which has been already exited. So for this atom, I will be providing one more extra energy that is emu. means that process has been not completed. So what is happening the atoms which are involving for the stimulated emission is the atoms which has been exited and the energy will be provided emu that is the extra energy. So it is directly proportional to the extra energy provided in the exited state and the extra energy will be provided for the atoms which are in the N2 level that is exited state. So if I want to remove this proportionality constant since the process has been not completed so that's why I will be taking the constant as B2 and this is also emission emission will be taken placements the atom should be coming from the direction of 2 to 1 so that's why it will be taken as 2 to1 E mu and n_sub_2 take it as equation number where b21 will be called asin Ain coefficient of stimulated emission.

[00:06:01]Now totally we will be having a three equations which will be representing a rate of induced absorption, spontaneous emission and stimulated emission. Now I will be taking a condition at thermal equilibrium.

[00:06:21]So at a thermal equilibrium what will be happening?

[00:06:25]Rate of absorption so will be always equal to rate of emission.

[00:06:38]So normally this will be happening. So since there is a absorption also taking place in our equation as well as the emission also taken place. So we'll be uh rearranging this one in such a way that rate of induced absorption that is the one absorption which we have absorption will be equal to since we have a two emission so we'll be taking a sum of this one rate of spontaneous emission plus rate of stimulated emission.

[00:07:24]So we'll be rearranging the equation according to the general equation that is at a thermal equilibrium. So take it as equation number four. Now apply equation 1 comma 2 and 3 in equation number four.

[00:07:45]Just apply this value in the equation number four. So what will be happening?

[00:07:50]So you know that rate of induced absorption what is the value? B12 E mu and N_sub_1. So will be equal to rate of spontaneous emission. This is the spontaneous emission where A 21 and N_sub_2 that is the equation plus rate of stimulated emission. The equation which is representing is B21 E mu N_sub_2. So we'll be writing the equation according to the requirement.

[00:08:23]So next what will be happening just I want a energy density formula here. So that's why I will be shifting all for the right hand side. It will be B12 E mu n_sub_1 minus b21 e mu n_sub_2 is equal to a21 n_sub_2 just we will be shifting the equation now I will be taking energy density e mu as a common term outside and the equation will be going as b1 to n_sub_1 1 - B 21 N_sub_2 is equal to A21 end just I will be rearranging the equation according to our equation and now I want only the equation as energy density I don't want entire equation so I will be shifting this bracket term to the denominator minator the value will become a 21 n_sub_2 divided by b12 ns1 - b21 in now the same equation so since there is a n1 n2 term in the denominator I will be taking a n_sub_2 term in this uh denominator common the value will be n_sub_2 2 and since in the first term there is no n_sub_2 so what we need to do we need to divide the number what we have took it as a common minus b21 since there is a n1 so that will be coming as a common term outside and we will be canceling n_sub_2 n2 now now we will be rewriting the equation which has been already cancelled And once again what I will be doing I will be taking one more common. What is the common? So I will be taking B12 outside.

[00:10:40]B12 it's nothing but the Einstein coefficient for the induced absorption in the denominator. N1 divided by N_sub_2 minus B21 divided by B12 just I will be taking out right. So as soon as this is taken outside so you know that B12 is taken outside and here there is no B12 term that's why I have divided here. Now by using Boltzman formula.

[00:11:21]So what I will be doing I will be using a Boltsman formula. What is the Boltsman formula? States if I want to know the ratio of the atoms which are in the number of atoms in the exited state with respect to number of atoms in the ground state will be always equal to E to ^ of minus of delta E / KT. So but in my equation take it as equation number five.

[00:11:52]So what is happening in my equation?

[00:11:55]There is the value as n_sub_1 / n_sub_2.

[00:11:59]So if I rearrange this equation, the equation will become as n_sub_1 / n_sub_2 is equal to e ^ of plus. If I rearrange this, it will be going for the denominator. So 1 / e ^ of x can be written as e ^ of x. So the value will become plus delta E / KT. Take it as equation number six.

[00:12:25]Applying Applying equation applying equation six in equation five. So what will be happening? I will be applying n_sub_1 divided by n_sub_2 value as e ^ of delta e / kt. So now the energy density formula will become a 211 divided by b12 and n1 divided by n_sub_2 will be e ^ of delta e / kt minus b21 / b12 just I will rearrange this equation take it as equation number seven now we know the plank's formula for the energy density already emu already we know for the normal interaction of light the planks has already given a formula for this one so what we will do so we'll be writing that formula what is the formula e mu is equal to 8<unk>i h mu cq Q divided by CQ into 1 / E ^ of delta E / KT minus 1. This is the formula of energy density which has been given in general for the normal light for the induced absorption and spontaneous emission which will be taken place for the light. Now what we will do we will compare we'll compare equation seven and equation 8 we'll be comparing this equation what will be the comparison just observe in the place of the ratio in the place of a 21 divided by b12 what is there the value is 8 pi h mu cq divided by cq is there. So this is the comparison and one more comparison just observe. So B21 divided by B12 in the place now in the place of B21 divided by B12 what is there? The value is equated to 1. Now by comparing I can say that the rate of whatever the coefficient Einstein coefficient of stimulated emission will be equal to Einstein coefficient of induced absorption at a thermal equilibrium I can equate. So by generalizing the equation as soon as this is finished I can easily say that by comparing this one the energy density for the laser e mu is equal to a divided by b in spite of writing a 21 and b12 because nothing no term is there extra like in the b value now I can write it as a by b into e ^ of delta e divided by KT minus 1. So this formula will be representing the energy density formula for laser.

[00:16:02]So this is the actual derivation. So in the exam point of view this will be asked for the