# Classical/Relativistic electron through a potential difference

1. Mar 5, 2013

### Hakkinen

1. The problem statement, all variables and given/known data
An electron is accelerated through a potential difference of 2.044 MV. (a) find the kinetic energy, energy, speed (in c), and momentum classically. (b) Do the same relativistically.

Charge of the electron is $e = 1.6 \times 10^-19 C$

Rest mass is $m_o = 9.1 \times 10^-31 kg$

Rest energy is $E_o = 0.511 MeV$

2. Relevant equations

Classical equations
$u = c\sqrt{\frac{2eV}{E_o}}$

$T = qV = \frac{1}{2}mv^2$

$p = mv$

$E = T+U$ ???

Relativistic equations
$T = qV$

$p = γm_0v$

$E = T+E_0 = γm_0c^2$

3. The attempt at a solution

My only question with this problem is calculating the energy in each method.
Classically the electron's speed is calculated as approx 2.83c, so simply substituting in the known quantities you have:

$T \approx 2.06 MeV$

$p \approx 1.6 \times 10^-11 \frac{eV}{c}$

Now the total energy the electron possesses classically is just the sum of the potential and kinetic energies correct?

$E \approx 4.104 MeV ??$

Now relativistically:

Kinetic energy can be calculated with the same formula $T = qV \approx 2.06 MeV$

I solved for γ in $E = T + E_0$

$γ = \frac{T}{E_0} +1 = \frac{E}{E_0}$

then $β = \sqrt{1-\frac{E_0^2}{E^2}} \approx 0.98c$

Now with the velocity we can find relativistic momentum and energy

$p \approx 2.8 \times 10^-11 \frac{eV}{c}$

Now the relativistic energy is $E = T + E_0$ , is potential energy not included in the total relativistic energy? If the reasoning in the classical case is right, the electron would have more energy than in the relativistic case as the potential energy of 2.044 MeV would be added to it.

Relativistic E without U $E \approx 2.571 MeV$
with U $E \approx 4.61 MeV$

2. Mar 5, 2013

### Hakkinen

Hang on, I can't use $T = qV$ in the relativistic case correct?

3. Mar 5, 2013

### rude man

Classically, K = 1/2 mv^2, m = m0.
Relativistically, K = (m - m0)c2, m = γm0.

BTW the relativistic formula reduces to 1/2 m0v2 for v << c so it's really a general formula, as is of course required by the correspondence principle.

When a charged particle is propelled across a potential difference of V volts its K increases by qV. So 1 eV = K.E. gained by running an electron over a potential of 1V. Again, true classically and relativistically.

Just never assume K = 1/2 mv2 relativistically, even when you use the relativistic mass for m.

4. Mar 5, 2013

### Hakkinen

Thanks for the reply, I can't believe I didn't see think to use the relativistic KINETIC ENERGY equation for KE. I am quite sure I have the correct solutions now.

From $T = qV = γ(E_0 - 1)$ I solved for gamma $γ = \frac{qV}{E_0} + 1$

γ is about 5 so from there it was easy to correct the solutions and check that the relativistic KE formula gives approximately the same result as $T = qV$

5. Mar 7, 2013

### Hakkinen

Obviously I meant $T = E_0(γ - 1)$

6. Mar 7, 2013

Big 10-4!