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
An example of De Broglie equaton.........................
Explanation:
The De Broglie equation states that the
Another equation of De Broglie's work is
Where
So substitute the above equation in the below equation
As electrons mass is
Convert nm to m
650nm =
Plug in the variables
But you can't solve this equation so
Since we have calculated the momentum which is
When the momentum squares the expression changes
Recall that
It make's sense since the wavelength is so large .
#K = 5.7 xx 10^(-25)# #"J"#
Consider the properties of electrons... Since electrons have mass
#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
#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
#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...