Photoelectric effect: A. What is the maximum kinetic energy of electrons ejected from barium (W0=2.48eV) when illuminated by white li

Question

Photoelectric effect:
A. What is the maximum kinetic energy of electrons ejected from barium (W0=2.48eV) when illuminated by white light, lambda=410-750nm?
B. The work functions for sodium, cesium, copper, and iron are 2.3, 2.1, 4.7, and 4.5eV, respectively. Which of these metals will not emit electrons when visible light shines on it?

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niczorrrr 1 year 2021-09-01T09:07:13+00:00 1 Answers 0 views 0

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    2021-09-01T09:08:37+00:00

    Answer:

    A. K = 0.546 eV

    B. cooper and iron will not emit electrons

    Explanation:

    A. This is a problem about photoelectric effect. Then you have the following equation:

    [tex]K=h\nu-\Phi=h\frac{c}{\lambda} -\Phi[/tex]   (1)

    K: kinetic energy of the ejected electron

    Ф: Work function of the metal = 2.48eV

    h: Planck constant = 4.136*10^{-15} eV.s

    λ: wavelength of light = 410nm – 750nm

    c: speed of light = 3*10^8 m/s

    As you can see in the equation (1), higher the wavelength, lower the kinetic energy. Then, the maximum kinetic energy is obtained with the lower wavelength (410nm). Thus, you replace the values of all variables :

    [tex]K=(4.136*10^{-15}eV)\frac{3*10^8m/s}{410*10^{-9}m}-2.48eV\\\\K=0.546eV[/tex]

    B. First you calculate the energy of the photon with wavelengths of 410nm and 750nm

    [tex]E_1=(4.136*10^{-15}eV)\frac{3*10^{8}m/s}{410*10^{-9}m}=3.02eV\\\\E_2=(4.13610^{-15}eV)\frac{3*10^{8}m/s}{750*10^{-9}m}=1.6544eV[/tex]

    You compare the energies E1 and E2 with the work functions of the metals and you can conclude:

    sodium = 2.3eV < E1

    cesium = 2.1 eV < E1

    cooper = 4.7eV > E1 (this metal will not emit electrons)

    iron = 4.5eV > E1 (this metal will not emit electrons)

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