Archived How Do You Calculate the Wavelength of an Electron with 1keV Kinetic Energy?

[ZedCar]

Homework Statement

Homework Equations

de Broglie wavelength λ = h/p
E^2 = p^2 c^2 + (m0)^2 c^4
L-compton = (h-bar) / mc

The Attempt at a Solution

I'm trying to work out the wavelength of an electron with a kinetic energy of 1keV

So I intend on using
de Broglie wavelength λ = h/p

and

Relativistic mass equation
E^2 = p^2 c^2 + (m0)^2 c^4

then inputting the rest mass energy of an electron (511 keV) into the formula above. Though do I need to convert this first to kg?

Also, I'd still be left with two unknowns in mass eqn i.e. E and p, so how do I obtain the value for E?

Or should I instead be using the Compton wavelength formula i.e.

L-compton = (h-bar) / mc

The question just asks for the wavelength.

 

[QuantumQuest]

I'm trying to work out the wavelength of an electron with a kinetic energy of 1keV

The relationship between kinetic energy - momentum for a free electron is $T = \frac{p^{2}}{2m}$. So, from this, we can find the momentum $p$:

$p = \sqrt{2mT} = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 10^{3}\frac{kg^{2} m^{2}}{s^{2}}} \approx 17.064 \times 10^{-24} \frac{kg m}{s}$

Now, we can find the De Broglie wavelength: $\lambda = \frac{h}{p} = \frac{6.625 \times 10^{-34}}{17.064 \times 10^{-24}} m\approx 0.039 nm$

 

[Smalde]

The relationship between kinetic energy - momentum for a free electron is $T = \frac{p^{2}}{2m}$. So, from this, we can find the momentum $p$:

$p = \sqrt{2mT} = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 10^{3}\frac{kg^{2} m^{2}}{s^{2}}} \approx 17.064 \times 10^{-24} \frac{kg m}{s}$

Now, we can find the De Broglie wavelength: $\lambda = \frac{h}{p} = \frac{6.625 \times 10^{-34}}{17.064 \times 10^{-24}} m\approx 0.039 nm$

This result is correct, because 1keV is non-relativistic for electrons, since electron rest mass is around 500keV.

On the other hand one can solve this exercise using the equation OP wanted to use: E² = m²c⁴ + p²c²

He said the energy (meaning the kinetic energy) of the electron is one keV. Thus 1keV = E - mc² = $\sqrt{m²c⁴ + p²c²} -mc²$. If OP solves this equation, then OP can insert it in the De Broglie wavelength formula:

$$p² = \frac{1}{c²}(1keV + mc²)² - m²c² => p = \sqrt{ \frac{1}{c²}(1keV + mc²)² - m²c² } \approx 1.70933440*10^{-23}kg \frac{m}{s}$$

And so $\lambda = h/p \approx 0.038764033nm$

Both results, of course, agree

EDIT: Not trying to be pedant, but because I want to emphasize the minute differences between both methods I will recalculate what QuantumQuest already did, but with higher precision:

non-relativistic:

$p \approx 1.70849875*10^{-23}kg \frac{m}{s} \approx 31.9687keV/c$
$\lambda = h/p \approx 0.038782993nm$

relativistic:

$p \approx 1.70933440*10^{-23}kg \frac{m}{s} \approx 31.98434keV/c$
$\lambda = h/p \approx 0.038764033nm$

 

[Smalde]

Making my point about the minuteness of the corrections more thorough:

$p_{rel} = \sqrt{ \frac{1}{c²}(T + mc²)² - m²c² } = \sqrt{T²/c²+2Tm+m²c²-m²c²} = \sqrt{T²/c²+2Tm} = \sqrt{T}\sqrt{T/c²+2m}$
$p_{non-rel} = \sqrt{2Tm} = \sqrt{T}\sqrt{2m}$

Thus $\frac{p_{rel}}{p_{non-rel}} = \sqrt{ frac{T/c² + 2m}{2m}} = \sqrt{\frac{T}{2mc²} + 1}$
And so $\frac{p_{relativistic}}{p_{non-relativistic}}(x) = \sqrt{\frac{x}{2}+1}$ where $x = \frac{T}{mc²}$

This formula holds true for any massive particle.

In our case $x \approx 1/500$ and so the above formula is essentially one.
This is a good way to see when you can use non-relativistic approximations. Basically as long as your kinetic energies are much smaller than your rest mass energies, you are safe:

$$ T << mc² $$