Question 15.1: Find the kinetic energy of an electron with a de Broglie wav...
Find the kinetic energy of an electron with a de Broglie wavelength of 10 fm.
Using Eq. (13.14) of Chapter 13 and the de Broglie relation, the energy and the wavelength of an electron are seen to be related by the equation
E^{2} = p^{2}c^{2} + m^{2}c^{4}. (13.14)
E^{2} =\left( \frac{ hc} {λ}\right) ^{2} + m^{2}c^{4}.As we have discussed in the Introduction, the product of constants hc is equal to 1240 MeV · fm. Substituting this value of hc and λ = 10 fm into the above equation, we obtain
E = \sqrt{(124 MeV)^{2} + (0.511 MeV)^{2}} ≈ 124 MeV.The kinetic energy of an electron is equal to the difference between its energy and its rest energy of the electron
K.E. = 124 MeV − 0.511 MeV = 123.5 MeV.
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