#### Question

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.

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#### Solution

Temperature of the neutron, *T* = 300 K

Boltzmann constant, *k* = 1.38 × 10^{−23} kg m^{2} s^{−2} K^{−1}

Average kinetic energy of the neutron:

`K' = 3/2 kT`

`= 3/2 xx 1.38 xx 10^(-23) xx 300 = 6.21 xx 10^(-21) J`

The relation for the de Broglie wavelength is given as:

`lambda' = h/(sqrt(2K'm_n))`

Where

`m_n = 1.66 xx 10^(-27) Kg`

`h = 6.6 xx 10^(-34) Js`

`K' = 6.75 xx 10^(-21) J`

`:. lambda' = (6.63 xx 10^(-34))/sqrt(2xx 6.21 xx 10^(-21) xx 1.66 xx 10^(-27)) = 1.46 xx 10^(-10) m = 0.146 nm`

Therefore, the de Broglie wavelength of the neutron is 0.146 nm.

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#### Reference Material

Solution for question: 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. concept: Wave Nature of Matter. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science)