contestada

Derive an expression for the de Broglie wavelength of a thermal neutron and calculate its value at the temperature T = 290 K. The kinetic energy of the neutron at a given temperature T is approximately 3kT/2 where k = 1.38 x 10-23 J/K is the Boltzmann constant. The neutron mass is mn 1.67 x 10-27 kg and the Planck constant h = 6.626 x 10-34 Js.

Respuesta :

Answer:

1.479 A

Explanation:

Expression for de Broglie wave length

λ = [tex]\frac{h}{\left ( 2mKE \right )^{.5}}[/tex]

Where m is mass of neutron ,h is plank constant and KE is kinetic energy of

proton.

Kinetic Energy ( KE) = 3 / 2 kT

=1.5 X 1.38 X 10⁻²³ X 290 = 600.3 X 10⁻²³ J

[tex]\sqrt{2mKE}[/tex] = [tex]\sqrt{2\times1.67\times10^{-27}\times600.3\times10^{-23}}[/tex]

44.777 x 10⁻²⁵

λ = [tex]\frac{6.626\times10^{-34}}{44.777\times10^{-25}}[/tex]

=1.479 A.

Otras preguntas