Question

An electron has a de broglie wavelength equal to the diameter of a hydrogen atom in its groung state.

(a) What is the kinetic energy of the electron?
(b) How does the energy compare to the ground-state energy of the hydrogen atom?

Answer 1

(a)
`2.4\cdot 10^(-17) J`

The De Broglie wavelength of a particle is given by

`\lambda=(h)/(p)` (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

`K=(p^2)/(2m)`

where m is the mass of the particle. Re-arranging this equation,

`p=√(2mK)` (2)

And substituting (2) into (1),

`\lambda = (h)/(√(2mK))` (3)

For an electron,

`m=9.11\cdot 10^(-31)kg`

In the problem, the electron has a de broglie wavelength equal to the diameter of a hydrogen atom in the ground state:

`\lambda = d = 1\cdot 10^(-10) m`

So re-arranging eq.(3) we can find the kinetic energy of the electron:

`K=(h^2)/(2m\lambda^2)=((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`

(b) Approximately 10 times larger

The ground state energy of the hydrogen atom is

`E_0 = 13.6 eV`

Converting into Joules,

`E_0 =(13.6 eV)(1.6\cdot 10^(-19) J/eV)=2.2\cdot 10^(-18)J`

The kinetic energy of the electron in the previous part of the problem was

`E=2.4\cdot 10^(-17) J`

So, we see it is approximately 10 times larger.