Question
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800pm, calculate the characteristic velocity associated with the neutron.
✓
Answer
From de Broglie’s equation,
$\lambda=\frac{\text{h}}{\text{mv}}$
$\text{v}=\frac{\text{h}}{\text{m}\lambda}$
Where,
v = velocity of particle (neutron)
h = Planck’s constant
m = mass of particle (neutron)
$\lambda$ = wavelength
Substituting the values in the expression of velocity (v),
$\text{v}=\frac{6.626\times10^{-34}\text{Js}}{(1.67493\times10^{-27}\text{kg})(800\times10^{-12}\text{m})}$
$=4.94\times10^2\text{ms}^{-1}$
$\text{v}=494\text{ms}^{-1}$
$\therefore$ Velocity associated with the neutron = $494 \mathrm{ms}^{-1}$
$\lambda=\frac{\text{h}}{\text{mv}}$
$\text{v}=\frac{\text{h}}{\text{m}\lambda}$
Where,
v = velocity of particle (neutron)
h = Planck’s constant
m = mass of particle (neutron)
$\lambda$ = wavelength
Substituting the values in the expression of velocity (v),
$\text{v}=\frac{6.626\times10^{-34}\text{Js}}{(1.67493\times10^{-27}\text{kg})(800\times10^{-12}\text{m})}$
$=4.94\times10^2\text{ms}^{-1}$
$\text{v}=494\text{ms}^{-1}$
$\therefore$ Velocity associated with the neutron = $494 \mathrm{ms}^{-1}$
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