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The de-broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T and mass m is h/√(3mkT). Plz prove this?
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De-Broglie Wavelength of a Neutron in Thermal Equilibrium with Heavy Water

The de-Broglie wavelength is the wavelength associated with a moving particle. The wavelength is inversely proportional to the momentum of the particle. The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T and mass m is given by the equation:

λ = h/√(3mkT)

where λ is the de-Broglie wavelength, h is Planck's constant, k is Boltzmann's constant, and T is the temperature of the heavy water.

Derivation

The de-Broglie wavelength can be derived from the wave-particle duality principle, which states that matter can exhibit both wave and particle-like behavior. The wavelength of a particle is related to its momentum by the following equation:

λ = h/p

where p is the momentum of the particle.

The momentum of a neutron can be expressed in terms of its mass and velocity as follows:

p = mv

where m is the mass of the neutron and v is its velocity.

The velocity of a neutron in thermal equilibrium with heavy water can be expressed in terms of its kinetic energy as follows:

v = √(2E/m)

where E is the kinetic energy of the neutron.

The kinetic energy of a neutron in thermal equilibrium with heavy water can be expressed in terms of its temperature as follows:

E = 3/2 kT

Substituting these expressions into the equation for the de-Broglie wavelength gives:

λ = h/√(3mkT)

This is the desired result.

Conclusion

In conclusion, the de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T and mass m is given by the equation λ = h/√(3mkT). This equation can be derived from the wave-particle duality principle and the expressions for the momentum, velocity, and kinetic energy of a neutron in thermal equilibrium with heavy water.
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