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

(a) $v = 1.0\times 10^7~m/s$ (b) $v = 2.6\times 10^7~m/s$ (c) The particle is an alpha particle.
(a) We can convert the energy to units of joules: $300~eV \times \frac{1.6\times 10^{-19}~J}{1~eV} = 4.8\times 10^{-17}~J$ We can find the speed: $\frac{1}{2}mv^2 = 4.8\times 10^{-17}~J$ $v^2 = \frac{(2)(4.8\times 10^{-17}~J)}{m}$ $v = \sqrt{\frac{(2)(4.8\times 10^{-17}~J)}{m}}$ $v = \sqrt{\frac{(2)(4.8\times 10^{-17}~J)}{9.109\times 10^{-31}~kg}}$ $v = 1.0\times 10^7~m/s$ (b) We can convert the energy to units of joules: $3.5~MeV \times \frac{1.6\times 10^{-19}~J}{1~eV} = 5.6\times 10^{-13}~J$ We can find the speed: $\frac{1}{2}mv^2 = 5.6\times 10^{-13}~J$ $v^2 = \frac{(2)(5.6\times 10^{-13}~J)}{m}$ $v = \sqrt{\frac{(2)(5.6\times 10^{-13}~J)}{m}}$ $v = \sqrt{\frac{(2)(5.6\times 10^{-13}~J)}{1.67\times 10^{-27}~kg}}$ $v = 2.6\times 10^7~m/s$ (c) We can convert the energy to units of joules: $2.09~MeV \times \frac{1.6\times 10^{-19}~J}{1~eV} = 3.344\times 10^{-13}~J$ We can find the mass: $\frac{1}{2}mv^2 = 3.344\times 10^{-13}~J$ $m = \frac{(2)(3.344\times 10^{-13}~J)}{v^2}$ $m = \frac{(2)(3.344\times 10^{-13}~J)}{(1.0\times 10^{7}~m/s)^2}$ $m = 6.69\times 10^{-27}~kg$ We can see that this mass is very close to the mass of an alpha particle (2 protons and 2 neutrons), so we can assume that the particle is an alpha particle.