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

Md Ariful Islam

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Md Ariful Islam

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$lambda = "0.388 nm"$.

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 . 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.

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