Answer to Question #119947 in Quantum Mechanics for Dheeraj

Question

The de broglie wavelength, associated ith a neutron at teprature 27 degree celcius, is equals to? Also please tell me that i have seen some formulas of thermal wavelenth (in termas of 3mkT, 2mkT, mkT). which one is correct. i am confuse to use them.

Answer 1

The expression for the de Broglie wavelength associated with a neutron a temperature T is

$\lambda = \frac{h}{\sqrt{3mkT}}$

Here $h=6.626*10^{-34} J.s$ is planck's constant.

$m=1.67*10^{-27} kg$ is mass of neutron.

$k=1.38*10^{-23} J/K$ is Boltzmann constant.

$T=27^o=300 K$ is temperature.

On substituting all above values,

$\lambda = \frac{6.626*10^{-34} J.s}{\sqrt{3(1.67*10^{-27} kg)(1.38*10^{-23} J/K)(300 K)}}$

$=1.45*10^{-10} m =0.145 nm$