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A light source of wavelength λ illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 2.00 eV. A second light source with half the wavelength of the first ejects photoelectrons with a maximum kinetic energy of 6.00 eV. What is the work function of the metal?

Respuesta :

jorgezarate

Answer:

[tex]W=2eV[/tex]

Explanation:

Energy from a light source (photons):

E=h*c/λ

Photoelectric effect and Work function (W):

E=W+Ek

Ek: maximum kinetic energy from photoelectrons

then:

h*c/λ=W+Ek

Case 1:

[tex]h*c/lambda_{1}=W+Ek_{1}[/tex]  (1)

Case 2:

[tex]h*c/lambda_{2}=W+Ek_{2}[/tex]

but  [tex]lambda_{2}=lambda_{1}/2[/tex]

[tex]2h*c/lambda_{1}=W+Ek_{2}[/tex]  (2)

If we divide (2) by (1):

[tex]2=\frac{W+Ek_{2}}{W+Ek_{1}}[/tex]

[tex]W=Ek_{2}-2Ek{1}=2eV[/tex]

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