Answer:
A 0.914c
B. 5.13x10^8m/s
Explanation:
See attached file
Answer:
(a) the ratio of the speed of an electron, v having this energy to the speed of light is 0.914 c
(b) the speed of the electron is 5.136 x 10⁸ m/s
Explanation:
Given;
potential of the electron, V = 750 kV
kinetic energy of the electron, K.E = 7.50 * 10⁵ eV
Part (A) the ratio of the speed of an electron having this energy to the speed of light
E = k + mc²
where;
E is the total energy of the electron
k is kinetic energy
[tex]k = E -mc^2\\\\k = \frac{mc^2}{\sqrt{1-v^2/c^2} } -mc^2 = 7.5*10^5*1.6*10^{-19}\ J = 12*10^{-14} \ J\\\\ \frac{mc^2}{\sqrt{1-v^2/c^2} } =12*10^{-14} \ J + mc^2\\\\{\sqrt{1-v^2/c^2} = \frac{mc^2}{12*10^{-14} \ J + mc^2}} \\\\but, mass \ of \ electron, m = 9.1 *10^{-31} \ kg \ and \ speed \ of \ light\ c, =3*10^8 \ m/s\\\\{\sqrt{1-v^2/c^2} = \frac{9.1*10^{-31}(3*10^8)^2}{12*10^{-14} + \ 9.1*10^{-31}(3*10^8)^2}} = 0.4057\\\\1 - v^2/c^2 = 0.4057^2\\\\ 1 - v^2/c^2 = 0.1646\\\\v^2/c^2 = 1-0.1646\\\\[/tex]
[tex]v^2/c^2 = 0.8354\\\\\frac{v}{c} = \sqrt{0.8354}\\\\\frac{v}{c} = 0.914\\\\v = 0.914 \ c[/tex]
Part (B) speed of the electron when computed from principles of classical mechanics
[tex]k = \frac{1}{2}mv^2\\\\ v^2 = \frac{2k}{m} \\\\v = \sqrt{\frac{2k}{m} } \\\\v = \sqrt{\frac{2*12*10^{-14}}{9.1*10^{-31}} }\\\\v = 5.136*10^8 \ m/s[/tex]
