Question 13.1: Find the kinetic energy of an electron moving with velocity ...
Find the kinetic energy of an electron moving with velocity (a) v = 1.00 × 10^{−4} c, (b) v = 0.9 c.
(a) Since the velocity v = 1.00 × 10^{−4} c is far less than 11.5 percent the speed of light, we use the classical expression for the kinetic energy. We may write
KE =\frac{1}{2} mv²= \frac{1}{2} mc²(\frac{v}{c} )^{2}=\frac{1}{2}mc^{2}(1.00\times 10^{-4})^{2}.
Here the term mc² is the rest energy of the electron. Using the value 0.511 MeV given in Appendix A, we obtain
KE =2.56\times 10^{-3} eV.
| Appendix A | |||
| Constants and conversion factors | |||
| Constants | |||
| Speed of light | c | 2.99792458 × 10^{8} m/s | |
| Charge of electron | e | 1.6021773 × 10^{−19} C | |
| Plank’s constant | h | 6.626076 × 10^{−34} J s | |
| 4.135670 × 10^{−15} eV s | |||
| \hbar=h/2π | 1.054573 × 10^{−34} J s | ||
| 6.582122 × 10^{−16} eV s | |||
| hc | 1239.8424 eV nm | ||
| 1239.8424 MeV fm | |||
| Hydrogen ionization energy | 13.605698 eV | ||
| Rydberg constant | 1.0972 × 10^{5} cm^{−1} | ||
| Bohr radius | a_{0} = (4π \epsilon _{0})/(me²) | 5.2917725 × 10^{−11} m | |
| Bohr magneton | μ_{B} | 9.2740154 × 10^{−24} J/T | |
| 5.7883826 × 10^{−5} eV/T | |||
| Nuclear magneton | μ_{N} | 5.0507865 × 10^{−27} J/T | |
| 3.1524517 × 10^{−8} eV/T | |||
| Fine structure constant | α = e^{2}/(4π\epsilon _{0} c \hbar) | 1/137.035989 | |
| e^{2}/4π\epsilon _{0} | 1.439965 eV nm | ||
| Boltzmann constant | k | 1.38066 × 10^{−23} J/K | |
| 8.6174 × 10^{−5} eV/K | |||
| Avogadro’s constant | N_{A} | 6.022137 × 10^{23} mole | |
| Stefan-Boltzmann constant | σ | 5.6705 × 10^{−8} W/m² K^{4} | |
| Particle masses | |||
| kg | u | MeV/c² | |
| Electron | 9.1093897 × 10^{−31} | 5.485798 × 10^{−4} | 0.5109991 |
| Proton | 1.6726231 × 10^{−27} | 1.00727647 | 938.2723 |
| Neutron | 1.674955 × 10^{−27} | 1.008664924 | 939.5656 |
| Deuteron | 3.343586 × 10^{−27} | 2.013553 | 1875.6134 |
| Conversion factors | |||
| 1 eV | 1.6021773 × 10^{−19} J | ||
| 1 u | 931.4943 MeV/c² | ||
| 1.6605402 × 10^{−27} kg | |||
| 1 atomic unit | 27.2114 eV | ||
(b) Since the velocity v = 0.9 c is close to the speed of light, we use Eq. (13.13) in this case to obtain
KE = E − R = (γ − 1) mc², (13.13)
KE =\left(\frac{1}{\sqrt{1-(0.9)^{2}} }-1.0 \right) (0.511 \ MeV) = 0.661 MeV.
The kinetic energy of the electron is now greater than its rest energy.