Question

A particle cannot generally be localized to distances much smaller than its de Broglie wavelength. This fact can be taken to mean that a slow neutron appears to be larger to a target particle than does a fast neutron in the sense that the slow neutron has probabilities of being found over a larger volume of space. For a thermal neutron at room temperature of 300 K , find (b) the de Broglie wavelength.

Answer 1

The De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

How is the De broglie wavelength of a thermal neutron at room temperature calculated?

Temperature, T = 300K

Momentum, p = mv

Therefore v = p/m

Energy, E= 1/2 m( p/m) ²

Boltzman Energy= 3/2 KT

3/2KT = 1/2 m(p/m)²

Therefore p =  $\sqrt{3RTm}$

According to De broglie hypothesis, P = h ÷ λ

Therefore,    λ  = h ÷ $\sqrt{3RTm}$

                     = 6.6× $10^{-34}$ ÷  $\sqrt{3 \times 1.38 \times 10^{-23} \times 300 \times 1.6 \times 10^{-27} }$

                     = 0.15 × $10^{-9}$

Therefore the De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

To learn more about De broglie wavelength, refer: https://brainly.in/question/6131028

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