Question

What is the de Broglie wavelength of an electron that has been accelerated through a potential of `"10 V"`?

Answer 1

`lambda = "0.388 nm"`.

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Well, since an electron is a particle with mass, it can be described by the de Broglie relation:

`lambda = h/(mv)`

where:

• `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant. Remember that `"1 J" = ("1 kg"cdot"m"^2)/"s"`.
• `m_e = 9.109 xx 10^(-31) "kg"` is the rest mass of an electron.
• `v` is its speed. We don't need to know that per se.

And since it is also moving with a certain speed, it will have a kinetic energy of:

`K = 1/2 mv^2 = p^2/(2m)`

where:

• `p = mv` is the linear momentum of the particle.
• `m` is the mass of the particle.
• `v` is the speed of the particle.

And so, we can get the forward momentum in terms of the kinetic energy:

`p = sqrt(2mK) = mv`

Lastly, note that in `"1 eV"`, there is `1.602 xx 10^(-19)` `"J"`, i.e. one needs `"1 eV"` to push one electrons' worth of charge through a potential difference of `"1 V"` in an electric field.

Therefore, the de Broglie wavelength of the electron is:

`color(blue)(lambda) = h/sqrt(2m_eK_e)`

`= (6.626 xx 10^(-34) cancel"kg"cdot"m"^cancel(2)"/"cancel"s")/sqrt(2 cdot 9.109 xx 10^(-31) cancel"kg" cdot 10 cancel"eV" xx (1.602 xx 10^(-19) cancel"kg"cdotcancel("m"^2)"/"cancel("s"^2))/(cancel"1 eV"))`

`= 3.88 xx 10^(-10) "m"`

`=` `color(blue)("0.388 nm")`

And so, this is on the order of an X-ray photoelectron.