(a) [tex]2.4\cdot 10^{-17} J[/tex]
The De Broglie wavelength of a particle is given by
[tex]\lambda=\frac{h}{p}[/tex] (1)
where
h is the Planck constant
p is the momentum of the particle
We also know that the kinetic energy of a particle (K) is related to the momentum by the formula
[tex]K=\frac{p^2}{2m}[/tex]
where m is the mass of the particle. Re-arranging this equation,
[tex]p=\sqrt{2mK}[/tex] (2)
And substituting (2) into (1),
[tex]\lambda = \frac{h}{\sqrt{2mK}}[/tex] (3)
For an electron,
[tex]m=9.11\cdot 10^{-31}kg[/tex]
In the problem, the electron has a de broglie wavelength equal to the diameter of a hydrogen atom in the ground state:
[tex]\lambda = d = 1\cdot 10^{-10} m[/tex]
So re-arranging eq.(3) we can find the kinetic energy of the electron:
[tex]K=\frac{h^2}{2m\lambda^2}=\frac{(6.63\cdot 10^{-34}Js)^2}{2(9.11\cdot 10^{-31} kg)(1\cdot 10^{-10} m)^2}=2.4\cdot 10^{-17} J[/tex]
(b) Approximately 10 times larger
The ground state energy of the hydrogen atom is
[tex]E_0 = 13.6 eV[/tex]
Converting into Joules,
[tex]E_0 =(13.6 eV)(1.6\cdot 10^{-19} J/eV)=2.2\cdot 10^{-18}J[/tex]
The kinetic energy of the electron in the previous part of the problem was
[tex]E=2.4\cdot 10^{-17} J[/tex]
So, we see it is approximately 10 times larger.
