Answer
(a) $v = 1.0\times 10^7~m/s$
(b) $v = 2.6\times 10^7~m/s$
(c) alpha particle
Work Step by Step
(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.