The formula for kinetic energy is #E_("K") = frac(1)(2) m v^(2)#.
The rest mass of an electron is around #9.110 times 10^(- 31)# #"kg"#.
We are already provided with the amount of kinetic energy of the electron.
So let's find its velocity:
#Rightarrow 2.14 times 10^(- 22)# #"J"# #= frac(1)(2) times 9.110 times 10^(- 31)# #"kg"# #times v^(2)#
#Rightarrow 2.14 times 10^(- 22)# #"kg m"^(2)# #"s"^(- 2)# #= 4.555 times 10^(- 31)# #"kg"# #times v^(2)#
#Rightarrow 4.698 times 10^(8)# #"m"^(2)# #"s"^(- 2)# #= v^(2)#
#therefore v = 2.167 times 10^(4)# #"m s"^(- 1)#
Then, the formula for the de Broglie wavelength is #lambda = frac(h)(p)#.
Planck's constant is approximately #6.63 times 10^(- 34)# #"J s"#.
The momentum #p# can be written as #m v#, the product of mass and velocity.
#Rightarrow lambda = frac(h)(m v)#
Now, let's substitute the values of the electron's mass and velocity into the formula:
#Rightarrow lambda = frac(6.63 times 10^(- 34) " J s")(9.110 times 10^(- 31) " kg" times 2.167 times 10^(4) " m s"^(- 1))#
#Rightarrow lambda = frac(6.63 times 10^(- 34) " kg m"^(2) " s"^(- 1))(1.974 times 10^(- 26) " kg m s"^(- 1))#
#therefore lambda = 3.359 times 10^(- 8)# #"m"#
Therefore, the de Broglie wavelength of the electron is #3.359 times 10^(- 8)# #"m"#.
