State and Derive the Maxwell-Boltzmann's Canonical Distribution Law.
Derivation : Consider a system of particles in thermal equilibrium at a temperature \(T\), confined within a fixed volume \(V\). Let's assume this system consists of \(N\) particles with a total energy \(E\). Our goal is to find the distribution of energies among the particles using the principles of statistical mechanics.
To derive the Maxwell-Boltzmann's distribution, we will make use of the canonical ensemble, which assumes that the system is in contact with a heat reservoir at a constant temperature \(T\). This ensemble allows for energy fluctuations while keeping the temperature constant.
The probability (\(P(E_i)\)) of a particle having energy \(E_i\) is given by the Boltzmann factor:
\[P(E_i) = \frac{e^{-\beta E_i}}{Z}\]
In this equation :
- \(P(E_i)\) represents the probability of a particle having energy \(E_i\),
- \(\beta = \frac{1}{kT}\), where \(k\) is the Boltzmann constant,
- \(Z\) is the partition function, which ensures the probability distribution is properly normalized.
The partition function (\(Z\)) is defined as the sum of the Boltzmann factors over all possible states of the system:
\[Z = \sum_{i} e^{-\beta E_i}\]
Now, let's assume that the particles in the system are non-interacting, and their energies are solely kinetic. This assumption holds for an ideal gas.
The kinetic energy of a particle can be expressed as \(E_i = \frac{1}{2} m v_i^2\), where \(m\) is the mass of the particle and \(v_i\) is its velocity.
The volume element in momentum space is given by \(d^3p = p^2 \sin(\theta) dp d\theta d\phi\), where \(p\) represents the magnitude of momentum.
The number of states for a given momentum range is proportional to the volume element in momentum space. Therefore, the number of states within the momentum range (\(p, p+dp\)) can be expressed as:
\[g(p) dp = \frac{4\pi V}{h^3} p^2 dp\]
Here, \(h\) represents the Planck constant.
Since \(p = mv\), we can substitute this into the expression for the number of states, yielding :
\[g(v) dv = \frac{4\pi V}{h^3} (mv)^3 dv\]
By simplifying further, we obtain :
\[g(v) dv = \frac{4\pi V}{h^3} (mv)^3 dv\]
The density of states (\(g(E)\)) in terms of energy can be obtained by converting the expression for \(g(v)\) in terms of \(v\) to \(g(E)\) in terms of \(E\):
\[g(E) dE = g(v) dv = \frac{4\pi V}{h^3} (2m)^{3/2} \sqrt{E} dE\]
Now, let's substitute the expression for \(g(E)\) into the partition function \(Z\) :
\[Z = \int g(E) e^{-\beta E} dE = \int_0^\infty \frac{4\pi V}{h^3} (2m)^{3/2} \sqrt{E} e^{-\beta E} dE\]
The integral can be solved using calculus techniques, resulting in :
\[Z = \left(\frac{4\pi V}{h^3} (2m)^{3/2}\right) \left(\frac{1}{\beta}\right)^{5/2}\]
Finally, substituting the partition function \(Z\) back into the probability expression, we arrive at the Maxwell-Boltzmann distribution law:
\[P(E_i) = \frac{e^{-\beta E_i}}{Z} = \frac{1}{N} \left(\frac{m}{2\pi kT}\right)^{3/2} e^{-\frac{E_i}{kT}}\]
Here, \(N\) represents the total number of particles in the system.
Conclusion : The Maxwell-Boltzmann distribution law provides a statistical description of how energies are distributed among particles in a gas at a given temperature. It allows us to analyse and understand the behaviour of individual particles within a system and is crucial in various fields of physics. By deriving the Maxwell-Boltzmann distribution using simple language, we have gained insight into its underlying principles and its importance in statistical mechanics.
Also Read : Write Selection Rules for Beta Decay.
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