# 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 **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

#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...