# Find the cutoff wavelength for the pho toelectric

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Find the cutoff wavelength for the pho- toelectric effect. The speed of light is 3 × 10 8 m / s and Planck’s constant is
K = h ν - Φ = h c λ c - Φ = 0 .
martinez-rivera (dm42382) – Quest HW 20 – germain – (13622) 2 005(part2of2)10.0points Calculate the stopping potential if the inci- dent light has a wavelength of 223 nm. Correct answer: 1 . 36423 V. Explanation: Let : λ = 223 nm = 2 . 23 × 10 11 m . The stopping potential V is related to the kinetic energy of the photoelectrons by K = q e V , where q e is the charge on an electron. q e V = K = h c λ - Φ , V = h c q e λ - Φ q e = ( 6 . 626 × 10 - 34 J · s ) ( 3 × 10 8 m / s ) (1 . 602 × 10 - 19 C) (2 . 23 × 10 11 m) - 4 . 2 eV 1 . 602 × 10 - 19 C · 1 . 602 × 10 - 19 J 1 eV = 1 . 36423 V . 006 10.0points In the photoelectric effect, it is found that incident photons with 4 . 60 eV of energy will produce electrons with a maximum kinetic energy of 4 . 30 eV. What is the threshold frequency of this material? Planck’s constant is 6 . 62607 × 10 - 34 J · s . = (4 . 6 eV - 4 . 3 eV)(1 . 6 × 10 - 19 J / eV) 6 . 62607 × 10 - 34 J · s = 7 . 24411 × 10 13 Hz . 007 10.0points In the photoelectric effect, the maximum
Explanation: Let : E = 4 . 60 eV , K max = 4 . 30 eV , and h = 6 . 63 × 10 - 34 J · s . K max = h f - h f t = E - h f t h f t = E - K max f t = E - K max h