Question #459d4
1 Answer
Explanation:
Start by looking up the mass of an electron
#m_"electron" ~~ 9.1094 * 10^(-31)# #"kg"#
Now, according to the de Broglie hypothesis, matter can actually behave like a wave.
In this case, your electron will exhibit wave-like behavior and have an associated matter wave. The wavelength of this electron is called the de Broglie wavelength and is given by the equation
#color(blue)(ul(color(black)(lamda = h/p))) -># the de Broglie wavelength
Here
#p# is the momentum of the electron#lamda# is its de Broglie wavelength#h# is Planck's constant, equal to#6.626 * 10^(-34)"kg m"^2"s"^(-1)#
Notice that the equation uses the momentum of the electron, which depends on the velocity of the electron,
#color(blue)(ul(color(black)(p = m * v)))#
In your case, you have
#p = 9.1094 * 10^(-31)"kg" * 3 * 10^(7)"m s"^(-1)#
#p = 2.733 * 10^(-23)# #"kg m s"^(-1)#
Plug this into the equation that allows you to calculate the de Broglie wavelength to find
#lamda = (6.626 * 10^(-34) color(red)(cancel(color(black)("kg"))) "m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-1)))))/(2.733 * 10^(-23)color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))#
#color(darkgreen)(ul(color(black)(lamda = 2.4 * 10^(-11)color(white)(.)"m")))#
I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the velocity of the electron.