What is the de Broglie wavelength of an electron traveling at #2.0 * 10^8 m##/s#?

1 Answer
Diego Martínez Paz · Truong-Son N.
Jun 24, 2017

#lambda=3.64*10^-12# #"m"#

Explanation:

de Broglie wave equation#-># #lambda=h/p#

...where

  • #lambda# is the wavelength in #"m"#.

  • #p# (#"mass"(m)*"velocity"(v)#) is momentum
    (electron mass = #9.109*10^(-31)# #"kg"#).

  • #h# is Planck's constant #=6.626*10^-34 "J"("joule")*"s"("second")#.
    (1 Joule = #"1 kg"cdot"m"^2"/s"^2#)

Resolving...

#lambda=(6.626*10^-34 "J"cdot"s")/(mv)#

#lambda=(6.626*10^-34 "J"cdot"s")/((9.109*10^-31 "kg")(2.0*10^8 "m/s"))#

#lambda=(6.626*10^-34 "kg"cdot"m"^2"/s")/(18.2*10^-23 "kg"cdot"m/s")#

Here, everything is cancelled except #"m"#.

#lambda=3.64*10^-12# #"m"#

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