What is the kinetic energy for one electron whose wavelength is #"650 nm"#? Its rest mass is #9.109 xx 10^(-31) "kg"#.

2 Answers
May 6, 2017

An example of De Broglie equaton.........................

Explanation:

The De Broglie equation states that the

#"KE" = (1/2)"mv"^2#

#"Where KE is kinetic energy"#
#"Where m is mass and is a constant for each particle"#
#"Where v is velocity"#

#2"KE" = "mv"^2#

#2"KEm" = "m"^2"v"^2#

#"mv" = "p"#

#therefore "2KEm" = "p"^2#

#therefore p = sqrt("2KEm")#

Another equation of De Broglie's work is

#λ = "h"/"p"#

Where #"h"# is Planck's constant

So substitute the above equation in the below equation

#λ = "h"/ sqrt("2KEm")#

#sqrt("2KEm") = "h"/λ #

As electrons mass is #9.11 xx 10^(-31) "kg"#

Convert nm to m

650nm = #650xx10^-9m#

Plug in the variables

#sqrt("2KE"xx(9.11 xx 10^(-31) "kg")) = (6.626 xx 10^(-34) "kg m"^2 s^(-2) ⋅ s)/(650xx10^-9m) = #

#sqrt("2KE"xx(9.11 xx 10^(-31) "kg")) = 1.0193846xx10^(-27)" kg m⋅ s"^(-2) ⋅ "s"#

But you can't solve this equation so

#2"KE" = ("mv")^2#
#2"KEm" = p^2#

Since we have calculated the momentum which is #1.0193846xx10^(-27)" kg m⋅ s"^(-2) ⋅ "s"# KE is

#2KE xx (9.11 xx 10^(-31) "kg") = 1.0193846e-27^2#

When the momentum squares the expression changes

#2KE xx (9.11 xx 10^(-31) "kg") = 1.039145xx10^(-54)" kg"^2 "m"^2 s^(-2) ⋅ s#

#2"KE" = (1.039145xx10^(-54)" kg"^cancel(2) "m"^2 s^(-2) ⋅ s)/(9.11 xx 10^(-31) cancel("kg"))#

#2KE = 1.1406641xx10^(-24)" kg m"^2 "s"^(-2)#

#KE = (1.1406641xx 10^(-24)"kg m"^2 "s"^(-2))/2#

#KE = 5.7033205xx10^(-25)" kg m"^2 "s"^(-2)#

Recall that #"kg m"^2 "s"^(-2)# is equat to joules

#KE = 5.7033205xx10^(-25) J#

It make's sense since the wavelength is so large .

May 6, 2017

#K = 5.7 xx 10^(-25)# #"J"#


Consider the properties of electrons... Since electrons have mass #m# and a wavelength #lambda#, they follow the de Broglie relation:

#lambda = h/p = h/(mv)# #" "bb((1))#

where #h = 6.626 xx 10^(-34) "J"cdot"s"# is Planck's constant and #p = mv# is the linear momentum.

The kinetic energy can be expressed as a function of the momentum:

#K = 1/2 mv^2 = p^2/(2m)##" "bb((2))#

Therefore, if we solve #(1)# for momentum, we can get the kinetic energy from #(2)#.

#p = h/lambda#

#=> K = (h"/"lambda)^2/(2m)#

#= ((6.626 xx 10^(-34) "kg"cdot"m"^(cancel(2))"/s")/(650 cancel"nm" xx cancel"1 m"/(10^9 cancel"nm")))^2xx1/(2*9.109 xx 10^(-31) cancel"kg")#

Since the units work out as #"kg"cdot"m"^2"/s"^2#, or #"N"cdot"m"#, or #"J"#, we have units of energy. Therefore, the kinetic energy is:

#color(blue)(K = 5.7 xx 10^(-25))# #color(blue)("J")#

to two sig figs.

It makes sense that the kinetic energy is very small; after all, the electron is very light...