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) k T at 300 K.

Answer 1

a)

Given: The de Broglie wavelength of neutron is 1.4× 10 −10  m.

The de Broglie wavelength is given as,

λ= h m n v

Where, h is Planck's constant, m n is the mass of neutron and v is the velocity of neutron.

The Kinetic energy of neutron is given as,

K= 1 2 m n v 2 = 1 2 m n ( h m n λ ) 2 = 1 2 × h 2 m n λ 2

By substituting the values in the above equation, we get

K= ( 6.63× 10 −34 ) 2 2× ( 1.40× 10 −10 ) 2 ×1.675× 10 −27 = 6.7× 10 −21 1.6× 10 −19  eV =4.19× 10 −2  eV

Thus, the kinetic energy of the neutron is 4.19× 10 −2  eV.

b)

Given: The absolute temperature is 300 K.

The kinetic energy associated with the temperature is given as,

K= 3 2 kT

Where, k is the Boltzmann constant and T is the absolute temperature.

By substituting the given values in the above equation, we get

K= 3 2 ×1.38× 10 −23 ×300 =6.21× 10 −21  J

The de Broglie wavelength associated with this kinetic energy is given as,

λ= h 2 m n K

By substituting the values in the above equation, we get

λ= 6.63× 10 −34 2×1.675× 10 −27 ×6.21× 10 −21 =1.45× 10 −10 m

Thus, the de Broglie wavelength of neutron is 1.45× 10 −10 m.