What is the kinetic energy of each electron in a beam of electrons if the beam produces a diffraction pattern of a crystal which is similar to that of a beam of 1.00 eV neutrons? Note: The electron mass is 9.11 × 10^-³¹ kg;and the neutron mass is 1.67 × 10^-²⁷ kg.

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Quizplus · Accepted Answer

The de Broglie wavelength of a particle is given by:$$\lambda = \frac{h}{p}$$where h is Planck's constant and p is the momentum of the particle. The kinetic energy of a particle is given by:$$K = \frac{p^2}{2m}$$where m is the mass of the particle. We can combine these two equations to get:$$\lambda = \frac{h}{\sqrt{2mK}}$$If the diffraction pattern of a beam of electrons is similar to that of a beam of 1.00 eV neutrons, then the de Broglie wavelength of the electrons must be the same as the de Broglie wavelength of the neutrons. Therefore, we can set the two equations equal to each other and solve for K:$$\frac{h}{\sqrt{2m_eK}} = \frac{h}{\sqrt{2m_n(1.00 	ext{ eV})}}$$where me is the mass of the electron and mn is the mass of the neutron. Solving for K, we get:$$K = \frac{m_e}{m_n}(1.00 	ext{ eV}) = (9.11 	imes 10^{-31} 	ext{ kg})/(1.67 	imes 10^{-27} 	ext{ kg})(1.00 	ext{ eV}) = 1830 	ext{ eV}$$Therefore, the kinetic energy of each electron in the beam is 1830 eV.

What Is the Kinetic Energy of Each Electron in a Beam