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# Special relativity and the accelerated electron momentum

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What is an electron's momentum, if it is
accelerated across a 6 mm potential difference
of 15.6 mV? Answer in units of keV/c.

(Given: The charge on the electron is qe = 1.60218 *10 ^-19 C.
Given: The mass of the electron is me = 9.10939 * 10 ^-31 kg)

© BrainMass Inc. brainmass.com October 5, 2022, 2:40 am ad1c9bdddf
https://brainmass.com/physics/special-relativity/special-relativity-and-the-accelerated-electron-momentum-145796

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When accelerated by a potential difference of V, electron gains a kinetic energy of eV.

This time since the accelerating voltage is very small, electrons kinetic energy is very small as well. We can use the non relativistic expression for kinetic energy.

K = p^2/2m
eV = p^2/2m
Multiply numerator and denominator with c^2
eV = p^2c^2/2mc^2
p^2c^2 = eV * 2 mc^2

eV = 0.0156 eV = 0.0156 x 10^-3 keV
Electron Rest energy (mc^2) = 0.51 Mev = 510 keV
Hence
p^2c^2 = eV * 2 mc^2 = 0.0156 x 10^-3 * 2 * 510 =
pc = 0.126 keV
p = 0.126 keV/c

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If you use the relativistic expressions like last time you will get the same result as follows:

Upon acceleration, electron gains a kinetic energy (K) of eV where V is the accelerating voltage and e is electronic charge.
K = eV
Total energy E = K + moc^2 = eV + moc^2
But
E2 = p2c2 + (moc2)2
(eV + moc^2)2 = p2c2 + (moc2)2
(eV)2 + 2 eV moc2 = p2c2

pc = sqrt[(eV)2 + 2 eV moc2]

pc = sqrt[(1.6x10^-19*0.0156)2 + 2 *1.6x10^-19*0.0156 *
9.10939X10^-31*(3x10^8)2]

pc = 2.0219 x 10^-17 J = 2.0219 x 10^-17 /1.6x10^-19 eV =0.126 keV
p = 0.126 keV/c

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com October 5, 2022, 2:40 am ad1c9bdddf>
https://brainmass.com/physics/special-relativity/special-relativity-and-the-accelerated-electron-momentum-145796