731. Problem 50.19 (RHK) A beam of low-energy neutrons emerges from a reactor and is diffracted from a crystal. The kinetic energy of the neutrons are contained in a band of width centred on the kinetic energy K. We have to show that the angles for a given order of diffraction are spread over a range given in degrees by
Where is the diffraction angle for a neutron with kinetic energy K. |
Solution: Click For PDF Version The de Broglie wavelength of a neutron of kinetic energy K is given by the expression , as momentum of a neutron of mass is . Therefore, as the spread of kinetic energy is centred on the kinetic energy K the corresponding spread of de Broglie wavelengths will be . This is obtained by differentiating the de Broglie wavelength with respect to the variable K. If the grating separation is d , the grating equation for diffraction maximum of order at diffraction angle will be . Therefore, the spread of diffraction angle if the spread of wavelength is will be |