consider a system of N non-interecting particles in equation. These are 2 energy level E1 has N1 particles and E2 has N2 point N1 are oriented parallel to B and N2 are anti-parallel to B. Now plot the following parameters with T 1. Partition Function Z. 2. Occupation number of each level. 3. Magnetisation. 4. Internal energy 5. Specific Heat 6. Entropy

AIM:- consider a system of N non-interecting particles in equation. These are 2 energy level E1 has N1 particles and E2 has N2 point N1 are oriented parallel to B and N2 are anti-parallel to B. Now plot the following parameters with T

1. Partition Function Z.2. Occupation number of each level. 3. Magnetisation.4. Internal energy5. Specific Heat6. Entropy

THEORY AND FORMULA USED:-

   For Partition Function:- For Occupation Number:-
For Magnetisation:-
Z = (2*cosh((μ*B)/(K*T))
N = (exp((μ*B)/(K*T)))/(2*cosh((μ*B)/(K*T)))
M/N = μ*tanh((μ*B)/(K*T))
   For Specific heat at constant applied magnetic field:-
CB = (K*((( μB)./(K*T)).^2)(sech((μ*B)./(K*T))).^2)
 For Internal Energy:- For Entropy:-
 U = - μ*B*tanh((μ*B)/(K*T))
S = K*(log(2*cosh((μ*B)/(K*T)))-(( μ*B)/(K*T))*tanh((μ*B)/(K*T)))
 Here ,K = Boltzmann’s Constant B = Magnetic field
μ = Chemical Potential
T = Absolute temperature

ALGORITHM:-

1. Write the value of all given constant k,B,μ etc .
2. Write the formula of E1 and E2 energy .
3. Using the linspace command for define the range .
4. Using the ‘for loop’ for no. of points .
5. Write the equation of partition function, occupation number, magnetization, internal energy, specific heat and entropy .
6. Using the ‘scf’ command for set the graphic window .
7. Plot the graph b/w temp. v/s partition function, occupation number, magnetasation, internal energy and entropy .
8. Leveling the graph and give the title on the graph .and using ‘xgrid’ command.

 PROGRAM:-

clc
clf
k=8.617e-5
U=5.78e-5
B=1
E1=-U*B
E2=U*B t=linspace(1,10,1000) T=linspace(0.1,10,1000) for i=1:1000
u(i)=k*t(i)
b(i)=1/u(i) z(i)=2*cosh((U*B)/u(i)) x(i)=(1/z(i))*exp(-b(i)*E1) y(i)=(1/z(i))*exp(-b(i)*E2) m(i)=U*tanh((U*B)/k*t(i))
  //ev/k //ev/tesla
//tesla
  // x=N1/N //y=N1/N
// magnetisation

g(i)=-m(i) //internal energy
cb=(((U*B/T(i))^2)/k)*(sech(U*B/(k*T(i))))^2 // specific heat
 s(i)=-B*U/T(i)*tanh(U*B)/(k*T(i))+k*log(2*cosh(U*B/(k*T(i)))) end
scf(0) //partition function plot(t,z,'r')
xlabel('temp-->','fontsize',3) ylabel('z-->','fontsize',3)
title('partition function','fontsize',3) xgrid
scf(1) // occupation number
plot(t,x,'b')
plot(t,y,'b')
xlabel('temp-->','fontsize',3)
ylabel('occupation no.','fontsize',3) title('OCCUPATION NUMBER vs temp','fontsize',3) xgrid
disp(x(i)) disp(y(i))
scf(2) // magnetisation plot(t,m,'g') xlabel('temp-->','fontsize',3) ylabel('magnetisation','fontsize',3) title('magnetisation vs temp','fontsize',3) xgrid
scf(3) //internal energy plot(t,g,'k')
xlabel('temp-->','fontsize',3) ylabel('internal energy','fontsize',3) title('internal energy vs temp','fontsize',3) xgrid
scf(4) // specific Heat plot(T,cb,'r') xlabel('temp-->','fontsize',3) ylabel('specific heat','fontsize',3)
// Entropy
                         
title('specific Heat vs temp','fontsize',3) xgrid
scf(5) //Entropy plot(t,s,'r') xlabel('temp-->','fontsize',3) ylabel('Entropy','fontsize',3) title('Entropy vs temp','fontsize',3) xgrid

RESULT:-

1. Occupation number:- For 0 Kelvin, all the particles are in ground state so occupation number tends to 1. But as we increase the temperature particle gains the energy and get excited so occupation number for the first excited state increases whereas it decreases for the ground state.
2. Magnetisation:- When a paramagnetic substance is placed in magnetic field, there are two factors which governs the orientation of dipoles which are magnetic field and the temperature. The field tends to align dipoles in its direction whereas temperature disorders them. So, for a constant applied magnetic field as the temperature is increased, the dipoles get disordered and follows the Curie’s Law.
3. Specific Heat:- Specific heat of two-level system is different that of from solids because there is an upper bound on the energy level of two-level system.
4. Internal Energy:- At 0 kelvin all the dipoles are oriented in the field direction and energy is negative (-μ.B). But as we increase the temperature dipole gets
disordered or antiparallel to the field, their energy is positive (+μ.B) and hence total internal energy tends to zero at higher temperature.
5. Entropy:- As the temperature is increased, system tends to equilibrium and hence the entropy is maximum.


curves:-