In a photoelectric effect experiment you illuminate a metal with light of an unknown wavelength and measure the maximum kinetic energy of th

Question

In a photoelectric effect experiment you illuminate a metal with light of an unknown wavelength and measure the maximum kinetic energy of the photoelectrons to be 0.65 eV. Then you illuminate the same metal with light of a wavelength known to be 2/3 of the first wavelength and measure a maximum kinetic energy of 1.9 eV for the photoelectrons. 1. Find the first wavelength, in nanometers. lambda_1 = 2. Find the metal’s work function, in electron volts.

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Gia Bảo 2 months 2021-08-22T11:37:51+00:00 1 Answers 0 views 0

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    2021-08-22T11:39:04+00:00

    Answer:

    1. λ₁ = 4.97 x 10⁻⁷ m = 497 nm

    2. ∅ = 3.96 x 10⁻¹⁹ J = 2.475 eV

    Explanation:

    Using Einstein’s Photoelectric Equation:

    Energy Given by Photon = Work Function + K.E of Electron

    hc/λ = ∅ + K.E

    where,

    h = Plank’s Constant = 6.63 x 10⁻³⁴ J.s

    c = speed of light = 3 x 10⁸ m/s

    λ = wavelength of light

    ∅ = work function

    K.E = Kinetic Energy of Electron

    FOR 1ST SCENARIO:

    K.E = 0.65 eV

    Therefore,

    hc/λ₁ = ∅ + 0.65 eV   ——- equation 1

    FOR 2ND SCENARIO:

    λ₂ = (2/3)λ₁

    K.E₂ = 1.9 eV

    Therefore,

    hc/(2/3)λ₁ = ∅ + 1.9 eV  

    (3/2)hc/λ₁ = ∅ + 1.9 eV   ——- equation 2

    1.

    Subtracting equation 1 from equation 2, we get:

    (3/2)hc/λ₁ – hc/λ₁ = ∅ + 1.9 eV – ∅ – 0.65 eV  

    (1/2)hc/λ₁ = 1.25 eV

    (1/2)(6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ₁ = (1.25 eV)(1.6 x 10⁻¹⁹ J/1 eV)

    λ₁ = (9.945 x 10⁻²⁶ J.m)/(2 x 10⁻¹⁹ J)

    λ₁ = 4.97 x 10⁻⁷ m = 497 nm

    2.

    Using values in equation 1:

    (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4.97 x 10⁻⁷ m) = ∅ + (0.65 eV)(1.6 x 10⁻¹⁹ J/1 eV)

    4 x 10⁻¹⁹ J – 1.04 x 10⁻¹⁹ J = ∅

    ∅ = 3.96 x 10⁻¹⁹ J = 2.475 eV

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