Answer
(a) $v = 3.66\times 10^7~m/s$
(b) $v = 2.68\times 10^7~m/s$
(c) The particle is an electron.
Work Step by Step
(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.