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# hwsol1 - The energy of one photon is E = hν =(6 626 ×...

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Physics 31 Spring, 2007 Solution to HW #1 Problem A A light source of wavelength λ illuminates a metal and ejects photoelectrons with a maximum kinetic en- ergy of 1.0 eV. A second light source with half the wavelength of the Frst ejects photoelectrons with a maximum kinetic en- ergy of 4.0 eV. What is the work function of the metal? We use the equation K max = φ = hc λ φ to set up two equations in the two unknowns λ and φ (the work function): 1eV= hc λ φ (1) 4eV= hc λ/ 2 φ =2 hc λ φ (2) Subtracting Eq. (1) from Eq (2) gives hc λ =3eV λ = hc 3eV = 1240 eV nm 3eV = 413 nm . Subtracting 2 × Eq. (1) from Eq. (2) gives φ =2 . 0eV . 2 –5 ±ind the energy of a 700 nm photon. E = hc λ = 1240 eV nm 700 nm =1 . 77 eV 2 –7 A 1.00 kW radio transmitter operates at a frequency of 880 kHz. How many photons per second does it emit?
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Unformatted text preview: The energy of one photon is E = hν = (6 . 626 × 10-34 J s) × (880 × 10 3 s-1 ) = 5 . 83 × 10-28 J / photon Use this value to convert the stated power in watts (Joules/second) to photons/second: 10 3 J / s 5 . 83 × 10-28 J / photon = 1 . 72 × 10 30 photons / s 2 –8 Under favorable circumstances the human eye can detect 1 . × 10 18 J of electromagnetic energy. How many 600 nm photons does this represent? It is simplest to get the photon energy Frst in Joules: E = hc λ = 1240 eV nm 600 nm = 2 . 07 eV = 2 . 07 eV × ( 1 . 602 × 10-19 J / eV ) = 3 . 31 × 10-19 J Now the number of photons is just 1 × 10-18 J 3 . 31 × 10-19 J ≈ 3 January 23, 2008...
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