AIM:- Plot the following functions with the energy at different temperatures
a) Maxwell-Boltzmann distributionb) Fermi-Dirac distributionc) Bose-Einstein distributiond) And compare all the three distributions for 100 Kelvin
Theory:-
1.Bose- Einstein statistics:- In case off Bose- particles are indistinguishable . So the interchange of two particle between two energy states will not produce any new state.
𝑛(𝑖) = g(i)/(exp{α + βε(i)} − 1)
2. Fermi-Dirac statistics:- In case of Fermi- Dirac Statistics applicable to particles like electrons and obeying Pauli exclusion principle , only one particle can occupy a single energy state.
𝑛(𝑖) = g(i)/(exp{α + βε(i)} + 1)
3. Maxwell-Boltzmann statistics:- The conditions in Maxwell- Boltzmann statistics are- (i) Particles are distinguishable i.e. there are no symmetry restrictions .
(ii) Each Eigen state Ith quantum group may Contain 0,1,2,3............n particles.
(iii) The total number of particles in the entire system is always constant i.e. n= n1+n2 +.....ni=∑ni = constant.
(iv) The sum of energies of all the particles in different quantum groups , taken together, constitutes the total energy of the system.
𝑛(𝑖) = g(i) /exp{α + βε(i)}
Algorithm – for u=kt –
1. Given the value of constant K and T .
2. Write the equation of U= KT .
3.Using the linespace command for define the range.
4.Using the for loop and write the equation of Fermi, Bose and Maxwell’s equation. 5. Using the subplot command and plot the graph of Fermi, Bose and Maxwell’s.
6. Leveling the level and title on the graph and using the legend command.
For u=o:-
1. Write the value of ‘K’ for different Statistics .
2. Gives the different temperature [T1, T2,T3,T4] .
3. Write the equation of U=0 .
4. Using the ‘ linspace command’ for define the range the range and for loop .
5. Write the equation of Maxwell-Boltzmann ,Fermi Dirac and Bose Einstein of different temperature [T1,T2,T3,T4] .
6. Using the subplot command and plot the graph of Maxwell ,Fermi and Bose Statistics of different temperature .
7. Leveling the level and title on the graph and using the legend command and xgrid command .
Program:- 2(a)
clc
clf
k=8.617e-5
t=100
u=k*t
b=1/u; r=linspace(0.01,5*u,100)' for i=1:100
f(i)=1/(exp(b*(r(i)-u))+1) e(i)=1/(exp(b*(r(i)-u))-1) v(i)=1/exp(b*(r(i)-u))
end
subplot(2,2,1)
plot(r,f,'r')
plot(r,e,'g')
plot(r,v,'b')
xlabel('Energy--->>','fontsize',3)
ylabel('occupation number'--->>','fontsize',3)
title('occupation number' v/s Energy for maxwell,bose and fermi statistics','fontsize',3)
Program:-2(b)
clc
clf
K1=0; //Maxwell-Boltzmann Statistics K2=1; //Fermi-Dirac Statistics K3=-1; //Bose-Einstien Statistics T1=1;
T2=5;
T3=10;
T4=20;
u=0;
x=linspace(1,50,100);
for i=1:100 v1(i)=1/(exp(x(i)/T1)+K1); v2(i)=1/(exp(x(i)/T2)+K1); v3(i)=1/(exp(x(i)/T3)+K1); v4(i)=1/(exp(x(i)/T4)+K1);
end
subplot(2,2,1)
plot(x,v1,'c')
plot(x,v2,'b')
plot(x,v3,'g')
plot(x,v4,'k')
xgrid
legend(['T1';'T2';'T3';'T4'])
xlabel('Energy','fontsize',3)
ylabel('occupation number'','fontsize',3)
title('occupation number' v/s Energy for M-B Statistics','fontsize',3)
y=linspace(-50,50,100);
for i=1:100 w1(i)=1/(exp(y(i)/T1)+K2); w2(i)=1/(exp(y(i)/T2)+K2); w3(i)=1/(exp(y(i)/T3)+K2); w4(i)=1/(exp(y(i)/T4)+K2); end
subplot(2,2,2)
plot(y,w1,'c')
plot(y,w2,'b')
plot(y,w3,'g')
plot(y,w4,'k')
xgrid
legend(['T1';'T2';'T3';'T4'])
xlabel('Energy','fontsize',3)
ylabel('occupation number'','fontsize',3)
title('occupation number' v/s Energy for F-D Statistics','fontsize',3)
x=linspace(1,50,100);
for i=1:100 m1(i)=1/(exp(x(i)/T1)+K3); m2(i)=1/(exp(x(i)/T2)+K3); m3(i)=1/(exp(x(i)/T3)+K3); m4(i)=1/(exp(x(i)/T4)+K3); end
subplot(2,2,3)
plot(x,m1,'c')
plot(x,m2,'b')
plot(x,m3,'g')
plot(x,m4,'k')
xgrid
legend(['T1';'T2';'T3';'T4'])
xlabel('Energy','fontsize',3)
ylabel('occupation number','fontsize',3)
title('occupation number v/s Energy for B-E Statistic','fontsize',3)
RESULT:-
According to the formula
ni = (gi/(exp β(Ei-μ)) + K)
We have taken chemical potential, = 0 so that our formula becomes ni = (gi/(exp (βEi)) + K)
• When K = -1, it is Bose Einstein Distribution. When energy is equal to chemical potential occupation number tends to infinity independent of temperature. As we increase the temperature, we observe net increase in occupation number corresponding to a particular energy state. This is because now the particles can gain the energy and excite thermally to upper states.
• When K = 0, it is Maxwell Boltzmann Distribution. As we increase the temperature, we observe net increase in occupation number corresponding to a particular energy state. This is because now the particles can gain the energy and excite thermally to upper states.
• When K = 1, it is Fermi Dirac Distribution. For this distribution the case is different as occupation number cannot exceed more than 1 due the Pauli’s exclusion principle. At 0 Kelvin the distribution function is a step potential and all energy states are filled below a particular energy level known as Fermi energy level and above which all states are empty. But as we increase the temperature, the average occupation number decreases below the Fermi level whereas it increases above the Fermi level. The average occupation number is always 1⁄2 at Fermi level which is independent of temperature.
