Respuesta :
The relationship of the photoelectric effect and the de Broglie expression allows us to find the result for the wavelength of the ejected electrons is:
- Wavelength of de Broglie is λ= 1.03 10⁻⁹m
The photoelectric effect was explained by Einstein assuming that the light rays behave like particles called photons, therefore the
[tex]E_{photon} = K + \Phi[/tex]
where [tex]E_{photon}[/tex] is the energy of the photon given by the Planck relation, K is the kinetic energy of the ejected electrons and Ф the work function of the material.
The Planck relationship states that the energy of the photons is proportional to the frequency.
[tex]E_{photon} = h f[/tex]
Where h is Planck's constant and f is the frequency of the photons.
They indicate The work function is Ф= 2.25 10⁻¹⁹ J, the energy of the photon [tex]E_{photon}[/tex] = 4.52 10⁻¹⁹ J, let's find the kinetic energy of the ejected electrons.
[tex]K = E_{photon} - \Phi[/tex]
Let's calculate.
K = (4.52-2.25) 10⁻¹⁹
K = 2.27 10⁻¹⁹ J
Kinetic energy is the energy of motion and is given by the relationship.
[tex]K = \frac{p^2}{2m}[/tex]
The wave-particle duality was established by de Broglie with the relation.
[tex]p = \frac{h}{\lambda }[/tex]
Let's replace.
[tex]K = \frac{1}{2m} (\frac{h}{\lambda} )^2 \\\lambda^2 = \frac{h^2}{2m K}[/tex]
let's calculate.
[tex]\lambda^2 = \frac{(6.63 \ 10^{-34})^2 }{2 \ 9.1 \ 10^{-31 } \ 2.27 \ 10^{-19}}[/tex]
[tex]\lambda = \sqrt{1.06397 \ 10^{-18}}[/tex]
λ = 1.03 10⁻⁹m
In conclusion with the relationship of the photoelectric effect and the de Broglie expression we can find the result for the wavelength of the ejected electrons is:
- Wavelength of de broglie is: λ = 1.03 10⁻⁹m
Learn more about the photoelectric effect here: brainly.com/question/25730863
