What is the de Broglie wavelength of an electron that has been accelerated through a potential of "10 V"10 V?
1 Answer
lambda = "0.388 nm"λ=0.388 nm .
Well, since an electron is a particle with mass, it can be described by the de Broglie relation:
lambda = h/(mv)λ=hmv where:
h = 6.626 xx 10^(-34) "J"cdot"s"h=6.626×10−34J⋅s is Planck's constant. Remember that"1 J" = ("1 kg"cdot"m"^2)/"s"1 J=1 kg⋅m2s .m_e = 9.109 xx 10^(-31) "kg"me=9.109×10−31kg is the rest mass of an electron.vv 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)K=12mv2=p22m where:
p = mvp=mv is the linear momentum of the particle.mm is the mass of the particle.vv is the speed of the particle.
And so, we can get the forward momentum in terms of the kinetic energy:
p = sqrt(2mK) = mvp=√2mK=mv
Lastly, note that in
Therefore, the de Broglie wavelength of the electron is:
color(blue)(lambda) = h/sqrt(2m_eK_e)λ=h√2meKe
= (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.
