Question

) 37.36 .. Electrons are accelerated through a potential difference of 750 kV, so that their kinetic energy is 7.50 * 105 eV. (a) What is the ratio of the speed v of an electron having this energy to the speed of light, c? (b) What would the speed be if it were computed from the principles of classical mechanics?

Answer 1

Answer:

A 0.914c

B. 5.13x10^8m/s

Explanation:

See attached file

Answer 2

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

$k = E -mc^2\\\\k = \frac{mc^2}{\sqrt{1-v^2/c^2} } -mc^2 = 7.5*10^5*1.6*10^{-19}\ J = 12*10^{-14} \ J\\\\ \frac{mc^2}{\sqrt{1-v^2/c^2} } =12*10^{-14} \ J + mc^2\\\\{\sqrt{1-v^2/c^2} = \frac{mc^2}{12*10^{-14} \ J + mc^2}} \\\\but, mass \ of \ electron, m = 9.1 *10^{-31} \ kg \ and \ speed \ of \ light\ c, =3*10^8 \ m/s\\\\{\sqrt{1-v^2/c^2} = \frac{9.1*10^{-31}(3*10^8)^2}{12*10^{-14} + \ 9.1*10^{-31}(3*10^8)^2}} = 0.4057\\\\1 - v^2/c^2 = 0.4057^2\\\\ 1 - v^2/c^2 = 0.1646\\\\v^2/c^2 = 1-0.1646$

$v^2/c^2 = 0.8354\\\\\frac{v}{c} = \sqrt{0.8354}\\\\\frac{v}{c} = 0.914\\\\v = 0.914 \ c$

Part (B) speed of the electron when computed from principles of classical mechanics

$k = \frac{1}{2}mv^2\\\\ v^2 = \frac{2k}{m} \\\\v = \sqrt{\frac{2k}{m} } \\\\v = \sqrt{\frac{2*12*10^{-14}}{9.1*10^{-31}} }\\\\v = 5.136*10^8 \ m/s$