Answer:
Part A:
The proton has a smaller wavelength than the electron.
[tex]\lambda_{proton} = 6.05x10^{-14}m[/tex] < [tex]\lambda_{electron} = 1.10x10^{-10}m[/tex]
Part B:
The proton has a smaller wavelength than the electron.
[tex]\lambda_{proton} = 1.29x10^{-13}m[/tex] < [tex]\lambda_{electron} = 5.525x10^{-12}m[/tex]
Explanation:
The wavelength of each particle can be determined by means of the De Broglie equation.
[tex]\lambda = \frac{h}{p}[/tex] (1)
Where h is the Planck's constant and p is the momentum.
[tex]\lambda = \frac{h}{mv}[/tex] (2)
Part A
Case for the electron:
[tex]\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}[/tex]
But [tex]J = Kg.m^{2}/s^{2}[/tex]
[tex]\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}[/tex]
[tex]\lambda = 1.10x10^{-10}m[/tex]
Case for the proton:
[tex]\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}[/tex]
[tex]\lambda = 6.05x10^{-14}m[/tex]
Hence, the proton has a smaller wavelength than the electron.
Part B
For part b, the wavelength of the electron and proton for that energy will be determined.
First, it is necessary to find the velocity associated to that kinetic energy:
[tex]KE = \frac{1}{2}mv^{2}[/tex]
[tex]2KE = mv^{2}[/tex]
[tex]v^{2} = \frac{2KE}{m}[/tex]
[tex]v = \sqrt{\frac{2KE}{m}}[/tex] (3)
Case for the electron:
[tex]v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}[/tex]
but [tex]1J = kg \cdot m^{2}/s^{2}[/tex]
[tex]v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}[/tex]
[tex]v = 1.316x10^{8}m/s[/tex]
Then, equation 2 can be used:
[tex]\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}[/tex]
[tex]\lambda = 5.525x10^{-12}m[/tex]
Case for the proton :
[tex]v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}[/tex]
But [tex]1J = kg \cdot m^{2}/s^{2}[/tex]
[tex]v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}[/tex]
[tex]v = 3.07x10^{6}m/s[/tex]
Then, equation 2 can be used:
[tex]\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}[/tex]
[tex]\lambda = 1.29x10^{-13}m[/tex]
Hence, the proton has a smaller wavelength than the electron.
