Question

(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−10 m? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) kT at 300 K.

Answer 1

(a) De-broglie's wavelength, λ = $\frac{h}{\sqrt{2mK}}$

or, K = h²/2mλ²

where h is Plank's constant i.e., hv= 6.63 × 10^-34 J.s

m is mass of neutron.i.e., m = 1.677 × 10^-27 Kg

λ is wavelength associated by neutron.i.e., λ = 1.4 × 10^-10 m

and K is kinetic energy of neutron.

so, K = (6.3 × 10^-34)²/(2 × 1.677 × 10^-27 × 1.4 × 10^-10)²

= 6.686 × 10^-21 J

we know, 1eV = 1.6 × 10^-19 J

so, K = 6.86 × 10^-21/(1.6 × 10^-19) eV

= 0.04178 eV

(b) K = 3/2 kT , where K is Boltzmann's constant and T is temperature.

so, De-broglie's wavelength, λ = h/√(2mK)

= h/√(3mkT)

[ note : don't confuse to see k and K, capital K shows kinetic energy and small k shows Boltzmann's constant]

here T = 300K, k = 1.38 × 10^-23 J/Kelvin

now, λ = (6.63 × 10^-34)/√(3 × 1.677 × 10^-27 × 1.38 × 10^-23 × 300)

= 6.63 × 10^-10/√(20.8)

= 6.63 × 10^-10/4.56

= 1.45 × 10^-10 m

= 1.45 A°

Answer 2

here's the answer and hope this helps you.