## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $v = 3.66\times 10^7~m/s$ (b) $v = 2.68\times 10^7~m/s$ (c) The particle is an electron.
(a) We can convert the energy to units of joules: $7.0~MeV \times \frac{1.6\times 10^{-19}~J}{1~eV} = 1.12\times 10^{-12}~J$ We can find the speed: $\frac{1}{2}mv^2 = 1.12\times 10^{-12}~J$ $v^2 = \frac{(2)(1.12\times 10^{-12}~J)}{m}$ $v = \sqrt{\frac{(2)(1.12\times 10^{-12}~J)}{m}}$ $v = \sqrt{\frac{(2)(1.12\times 10^{-12}~J)}{1.67\times 10^{-27}~kg}}$ $v = 3.66\times 10^7~m/s$ (b) We can convert the energy to units of joules: $15~MeV \times \frac{1.6\times 10^{-19}~J}{1~eV} = 2.4\times 10^{-12}~J$ We can find the speed: $\frac{1}{2}mv^2 = 2.4\times 10^{-12}~J$ $v^2 = \frac{(2)(2.4\times 10^{-12}~J)}{m}$ $v = \sqrt{\frac{(2)(2.4\times 10^{-12}~J)}{m}}$ $v = \sqrt{\frac{(2)(2.4\times 10^{-12}~J)}{(4)(1.67\times 10^{-27}~kg)}}$ $v = 2.68\times 10^7~m/s$ (c) We can convert the energy to units of joules: $1.14~keV \times \frac{1.6\times 10^{-19}~J}{1~eV} = 1.824\times 10^{-16}~J$ We can find the mass: $\frac{1}{2}mv^2 = 1.824\times 10^{-16}~J$ $m = \frac{(2)(1.824\times 10^{-16}~J)}{v^2}$ $m = \frac{(2)(1.824\times 10^{-16}~J)}{(2.0\times 10^{7}~m/s)^2}$ $m = 9.12\times 10^{-31}~kg$ We can see that this mass is very close to the mass of an electron, so we can assume that the particle is an electron.