Question

The de-Broglie wavelength associated with thermal neutron ( t=51C say) is of the order of the

• A. distance between the atoms in the crystal
• B. size of the nucleus
• C. Bohr's radius
• D. size of grain

Answer 1

The correct option is A distance between the atoms in the crystal

$T={51}^{\circ }C$

So, temp in Kelvin = $51+273=324K$

de-broglie wavelength $\lambda =\frac{h}{p}$

$p=\sqrt{2mE}$

$\lambda =\frac{h}{\sqrt{2mE}}$

$m$ of neutron $=1.67\times {10}^{-27}kg$

Also, $E=\frac{3}{2}kT$

where $k=\text{Boltzman constant}=1.38\times {10}^{-23}Jmo{l}^{-1}{K}^{-1}$

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

Substituting the values,

$\lambda =1.39\times {10}^{10}m$

which is the order of distance between atoms in the crystal.