# Special relativity and the accelerated electron momentum

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)

https://brainmass.com/physics/special-relativity/special-relativity-and-the-accelerated-electron-momentum-145796

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Answer:

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

https://brainmass.com/physics/special-relativity/special-relativity-and-the-accelerated-electron-momentum-145796