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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have an electron beam of this energy is suitable for crystal diffraction experiments. - Physics

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प्रश्न

Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn= 1.675 × 10−27 kg)

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उत्तर

De Broglie wavelength = 2.327 × 10−12 m; neutron is not suitable for the diffraction experiment.

Kinetic energy of the neutron, K = 150 eV

= 150 × 1.6 × 10−19

= 2.4 × 10−17 J

Mass of a neutron, mn = 1.675 × 10−27 kg

The kinetic energy of the neutron is given by the relation:

`"K" = 1/2 "m"_"n""v"^2`

`"m"_"n""v" = sqrt(2 "KM"_"n")`

Where

v = Velocity of the neutron

mnv = Momentum of the neutron

De-Broglie wavelength of the neutron is given as:

`lambda = "h"/("m"_"n""v") = "h"/(sqrt(2 "Km"_"n"))`

it is clear that wavelength is inversely proportional to the square rot of mass.

Hence wavelength decrease with an increase in mass and vice versa.

∴ `lambda = (6.6 xx 10^(-34))/sqrt(2 xx 2.4 xx 10^(-17) xx 1.675 xx 10^(-27))`

= 2.327 × 10−12 m

It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å, i.e., 10−10 m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, a neutron beam of energy 150 eV is not suitable for diffraction experiments.

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