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  • A neutron beam of energy E scatters from atoms on a surface with a spacing d = 0.1nm. The first maximum of intensity in the reflected beam occurs at theta = 30 degree. What is the kinetic energy E of the beam in eV?

A neutron beam of energy E scatters from atoms on a surface with a spacing d = 0.1nm. The first maximum of intensity in the reflected beam occurs at theta = 30 degree. What is the kinetic energy E of the beam in eV?

Answers (1)

According to the Bragg’s law of diffraction, condition for the nth maxima is given by

2d\sin \theta=n \lambda\\\\ n=1\\\\ so\; \lambda=2d\sin \theta [\theta=30^{\circ}]\\\\ =2*0.1*10^{-9}\sin 30^{\circ}\\\\ p=\frac{h}{\lambda}=6.6*\frac{10^{-34}}{10^{-10}}=6.6*10^{-24}kg\frac{m}{s}\\\\ E=\frac{1}{2}mv^{2}=\frac{\frac{1}{2}m^{2}v^{2}}{m}\\\\ =\frac{p^{2}}{2m}=\frac{6.6*6.6*10^{-48}}{2*1.6*10^{-27}*1.6*10^{-19}}eV\\ =\frac{66*66*10^{-48+46}}{2*16*16}=33*\frac{33}{126}*10^{-2}\\\\ =8.5*10^{-2}=0.085eV

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Safeer PP

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