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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5 months 2021-08-22T11:37:51+00:00 1 Answers 11 views 0

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