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Hand in 8 Solutions(07)

# Hand in 8 Solutions(07) - Solutions for Hand-In Set#8 PHYS...

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Solutions for Hand-In Set #8 PHYS 212 - Spring 2007 A54 — Phtotoelectric Effect Application of simple classical theories of the photo- electric effect suggests the following: The energy of emitted electrons should depend on the intensity of the light; this is contradicted by experiment. There should be an observable time delay for an electron to absorb enough energy to be emitted by low-intensity light; electrons are observed to be emitted “instantaneously.” A55 — Kinetic energies of emitted electrons This is a two-part problem; both parts use the equation governing the photoelectic effect: K max = hf - φ. The information about the results of the experiment with 700nm light allows us to determine the work function φ . Rearranging the equation above gives φ = hf - K max = hc λ - K max = 1240 eV · nm 700 nm - 0 . 25 eV = 1 . 77 eV - 0 . 25 eV = 1 . 52 eV . For the 400nm light we have K max = hf - φ = hc λ - φ = 1240 eV · nm 400 nm - 1 . 52 eV = 3 . 1 eV - 1 . 52 eV = 1 . 58 eV . A59 — Electron in a box (a) The relationship between kinetic energy and momen- tum is the same as it is in classical mechanics: K = p 2 2 m , so p = 2 mK = q 2 × 5 . 11 × 10 5 eV/ c 2 × 75 eV = 8 . 75 keV/ c, or equivalently p = 2 mK = p 2 × 9 . 1 × 10 - 31 kg × 75 eV × 1 . 6 × 10 - 19 J/eV = 4 . 67 × 10 - 24 kg · m/s . (b) Using the de Broglie relationship gives λ = h p = hc pc = 1240 eV · nm 8 . 75 × 10 3 eV = 0 . 142 nm or equivalently λ = h p = 6 . 63 × 10 - 34 J · s 4 . 67 × 10 - 24 kg · m/s = 1 . 42 × 10 - 10 m (c) Second Excited State λ b Low probability (d) From the sketch of the wavefunction it is clear that the box is 1.5 wavelengths long, or b = 1 . 5 × λ = 1 . 5 × 0 . 142 nm = 0 . 212 nm .

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A60 — Whirl-a-Tune and quantization
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Hand in 8 Solutions(07) - Solutions for Hand-In Set#8 PHYS...

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