Question

In a beam of electrons used in a diffraction experiment, each electron is accelerated to a kinetic energy of \(150 \mathrm{keV} .\) (a) Are the electrons relativistic? Explain. (b) How fast are the electrons moving?

Short answer

Calculate their velocity. Answer: The electrons are NOT relativistic, as the ratio of their kinetic energy to their rest mass energy is 0.293. Their velocity is approximately 2.998 × 10^8 m/s.

Step-by-step solution

Part (a): Determine if electrons are relativistic

To determine if the electrons are relativistic, we need to compare their kinetic energy with their rest mass energy. If the ratio is much greater than 1, then we can consider the electrons to be relativistic. The rest mass energy of an electron is given by the formula: \(E_0 = m_{e} c^2\) where \(m_{e}\) is the electron mass (\(m_{e} = 9.10938 × 10^{-31} \ kg\)), and \(c\) is the speed of light (\(c = 2.998 × 10^8 \ m/s\)). To make it easier to compare the kinetic energy, we can express the rest mass energy in keV by using the conversion factor \(1 \mathrm{keV} = 1.60218 × 10^{-16} \ J\). \(E_0 = m_{e} c^2 = 9.10938 × 10^{-31} \cdot (2.998 × 10^8)^2 = 8.1871 × 10^{-14} \ J\) Now we can convert this to keV: \(E_0 = 8.1871 × 10^{-14} \ J \cdot \frac{1 \mathrm{keV}}{1.60218 × 10^{-16} \ J} = 511 \mathrm{keV}\) Now we can calculate the ratio \(R\) of kinetic energy (\(150 \mathrm{keV}\)) to rest mass energy (\(511 \mathrm{keV}\)): \(R = \frac{150 \mathrm{keV}}{511 \mathrm{keV}} = 0.293\) Since this ratio is less than 1, we conclude that the electrons are NOT relativistic.

Part (b): Calculate the velocity of the electrons

Since we determined that the electrons are not relativistic, we can use the non-relativistic formula for the kinetic energy: \(K = \frac{1}{2} m_{e} v^2\) We know the kinetic energy is \(150 \mathrm{keV}\) and the electron mass is \(9.10938 × 10^{-31} \ kg\). We can convert the kinetic energy back to Joules: \(K = 150 \mathrm{keV} \cdot 1.60218 × 10^{-16} \ J/\mathrm{keV} = 2.40327 × 10^{-14} \ J\) Now we can solve for the velocity: \(v^2 = \frac{2K}{m_{e}} = \frac{2 \cdot 2.40327 × 10^{-14}}{9.10938 × 10^{-31}} = (2.998 × 10^8)^2\) \(v = \sqrt{(2.998 × 10^8)^2} = 2.998 × 10^8 \ m/s\) The electrons are moving at a velocity of approximately \(2.998 × 10^8 \ m/s\).