What is the de Broglie wavelength of an electron traveling at #2.0 * 10^8 m##/s#?
1 Answer
Jun 24, 2017
Explanation:
de Broglie wave equation
...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
#lambda=3.64*10^-12# #"m"#