A solid contains N magnetic atoms having spin 1 / 2. At sufficiently high temperatures each spin

A solid contains N magnetic atoms having spin 1 / 2. At...

Key Concepts

Phase Transitions and Magnetic Ordering

Magnetic solids, such as those exhibiting ferromagnetism, undergo a transition from a high-temperature disordered phase (random spin orientations) to a low-temperature ordered phase (aligned spins). This transition is accompanied by a significant reduction in entropy. Understanding this process is key in analyzing thermodynamic properties, such as the specific heat peak associated with the ordering transition.

Relation Between Specific Heat and Entropy

The specific heat of a system is linked to how its entropy changes with temperature. This is formalized by the relation dS/dT = C/T, meaning that integrating the specific heat divided by temperature over a temperature range yields the change in entropy. This connection is crucial for determining parameters like the maximum specific heat by matching the total entropy change across a transition.

Entropy of a Two?State System

In systems with spin 1/2 particles, each particle can exist in one of two states. At high temperatures the randomness leads to a maximum entropy, characterized by the expression N·kB·ln(2), stemming from the two equally likely configurations per spin. This fundamental concept quantifies the degree of disorder in a system with discrete microstates.

Transcript

00:03 In this question the energy levels are separated by this is upper energy level and this is low energy levels and they are separated by kb delta.

00:21 Here kb is e bolsman constant that means e1 minus e0 that is e0 is equal to kb delta and this type of the system is known the two an two level system and this type of the splitting is because of the magnetic field or any other regions in this question we have to find out the specific heat see okay in order to find out this specific heat we have to find out the energy for the system and then we will to first step number one find out the energy find out energy for this two level system to energy level system and suppose this this is you say this is you and step number two then find out the specifically that is del u by delta at constant this delta let's do the first step step number one that is energy of this two sea two energy level system u is equal to energy in the low energy system that is e0 minus a 0 by k bt plus e 1 e power minus e 1 b bt plus e 1 e power minus e 1 by k bt divided by e power minus e0 by k bt plus e power minus even by k bt okay now from here we can take common e power minus e0 by k bt is common and it is e0 plus e1 dot e.

03:20 Power minus it will be e1 minus e0 by k b t and similarly in the denominator we can take e power minus e not by k b t and it is one plus e power minus e one plus e power minus e one minus e0 by kb t and this four factor is cancelled now this is e0 plus e 1 e power minus and this factor is kb delta as giving the question so we can write this e power k b delta divided by kb t similarly in the denominator 1 plus e power minus k b delta divided by kb t and this kb to kb cancel now the remaining part is u is equal to e0 plus even e power minus delta by t upon 1 plus e power minus delta by t as we know that in the question e1 minus e0 is equal to kb delta sorry this is kb delta from here e 1 is equal to e0 plus k b delta so here this e e can be placed by this factor e0 plus k b delta.

05:57 That is e0 plus a 0 plus k b delta e power minus delta by t upon 1 plus e power minus delta by t here e0 e0 is common that means e0 and here 1 plus this factor is he common here that is e power minus delta by t one more term is that is kb delta and e power minus delta by t and here we can write it separately one plus e power minus delta by t this one plus e power minus delta by t and this two is cancel now the u is equal to e0 plus kb delta e power minus delta by t divided by 1 plus e power minus delta by t if i take this in the denominator then it will be u is equal to e0 plus k b delta upon this will be this one is one and this will be e power delta by t so this is the step number one that we have calculated the internal energy of this two level system now step number two find out this is heat that is c is equal to del u by del t at constant delta because this is constant now differentiate this equation equation number two with respect to temperature at constant delta here kb is also constant e node is also constant then c is equal to first is constant that is derivation of this is 0 and kb delta is also constant and this will be minus 1 plus e power delta by t whole square into e power delta by t into minus delta by t square now c is equal to k b delta into this minus minus plus and this delta by t square delta by t square into e power delta by t square into e power delta by t upon 1 plus e power delta by t full square now we can write this c is equal to k b delta by t full square and e power delta by t upon 1 plus e power delta by t upon 1 plus e power delta by t whole square this is the solution of part a in question that means we have to show that the specific heat is represented by this equation if we plot this capacity specific heat versus t by delta then you can see in the figure 11 this is t by delta and here there is a maxima and then it is decreed so such type of the behavior is known anomalies is known as the short key anomalies and you can see this value is around one this is c in units of kb and here you can see the value of c is very small is very small in this region here this is very small and here it is very small that me at a particular value of the t by delta it has the maximum specific heat that means when the t by delta is less than one it is very small and it is very small in the case of this in both cases the specific heat is very small and this type of the splitting is very useful in the rare earth...

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