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Assignment 3 - SOLUTIONS DUE ON SEPT 30, 2002 (MONDAY) 1:00 PM To be dropped off at my office (C884) 1. Which of the following scientist developed the uncertainty principle: (a) Bohr (b) Einstein (c) Heisenberg (d) Schrödinger Heisenberg 2. What is the name of the equation, H Ψ = E Ψ, and what is Ψ (psi) in this equation? The equation is the Schrödinger equation. Ψ is the symbol for a wavefunction. 3. (a) Calculate the energy of the process of an electron in a hydrogen atom going from the M shell (n = 3) to the K shell (n = 1), using Bohr’s energy expression (given below). Is the light emitted or absorbed, argue using the sign of ∆E. (will be marked) E = - R h c /n2 R = Rydberg constant = 1.097 × 107 m-1 h = Planck’s constant = 6.626 × 10-34 Js c = speed of light in a vacuum = 2.998 ×108 m s-1 ∆E = Efinal – Einitial = -Rhc (1/12 – 1/32) = -1.097 × 107 m-1 × 6.626 × 10-34 Js × 2.998 ×108 m s-1 ×(1/12 – 1/32) = - 1.937 × 10-18 J A photon of the energy - 1.937 × 10-18 J is emitted, since ∆E is negative (exothermic process). (b) Calculate the energy in the unit, kJ mol-1. (will be marked) molar energy: ∆E = - 1.937 × 10-18 J × 6.022 × 1023 mol-1 = -1166000 J mol-1 = -1166 kJ mol-1 (c) Calculate the frequency, ν, of the energy that is emitted or absorbed. The emitted/absorbed light belongs to which region of the electromagnetic spectrum. (will be marked) E = hν ν = E/h = 1.937 × 10-18 J/(6.626 × 10-34 Js) = 2.923 × 1015 s-1 = 2.923 × 1015 Hz The emitted light belongs to the UV region of the electromagnetic spectrum. 1 4. To cause a cesium atom on a metal surface to lose an electron, an energy of 2.0 × 102 kJ mol-1 is required. (a) Calculate the longest possible wavelength that can ionize a cesium atom. (b)What region of the electromagnetic spectrum is this radiation found? (c) If a wavelength of 400 nm is used, what is the kinetic energy of the photoelectron? (Ekin = m v2/2) (a) energy required for one atom instead of 1 mol atoms: 2.0 × 102 kJ mol-1/6.022 × 1023 mol-1 = 2.0 × 105 J mol-1/6.022 × 1023 mol-1 = 3.3 × 10-19 J E = hν ν = E/h = 3.3 × 10-19 J/(6.626 × 10-34 Js) = 5.0 × 1014 s-1 = 5.0 × 1014 Hz c = λν λ = c/ν = 2.998 × 108 m s-1/(5.0× 1014 s-1) = 6.0 × 10-7 m = 6.0 × 102× 10-9 m = 600 nm (b) This radiation is found in the visible region of the spectrum (orange). (c) c = λν ν = c/λ = 2.998 × 108 m s-1/(400× 10-9 m) = 7.50 × 1014 s-1 E = hν = 6.626 × 10-34 Js × 7.50 × 1014 s-1 = 4.97 × 10-19 J Ekin = E400nm - Eionization = 4.97 × 10-19 J - 3.3 × 10-19 J = 1.7 × 10-19 J Ekin = 1.7 × 10-19 J × 6.022 × 1023 mol-1 = 100000 J mol-1 = 1.0 × 102 kJ mol-1 5. (a) Calculate the wavelength (in nm) associated with a 1.0 × 102 g golf ball moving at 30 m/s. (b) How fast must the ball travel to have a wavelength of 5.6 × 10-3 nm? deBroglie equation: λ = h/mv (a) λ = h/mv = 6.626 × 10-34 Js/(1.0 × 102 g × 30 m s-1) = 6.626 × 10-34 Js/(0.1 kg × 30 m s-1) = 2.2 × 10-34 J s2 kg-1 m-1 = 2.2 × 10-34 kg m2 s-2 s2 kg-1 m-1 = 2.2 × 10-34 m (b) λ = h/mv v = h/m λ = 6.626 × 10-34 Js/(1.0 × 102 g × 5.6 × 10-3 nm) = 6.626 × 10-34 Js/(0.1 kg × 5.6 × 10-12 m) = 1.2 × 10-21 Js kg-1 m-1 = 1.2 × 10-21 kg m2 s-2 s kg-1 m-1 = 1.2 × 10-21 m s-1 6. A possible excited state for the H atom has an electron in a 5d orbital. List all possible sets of quantum numbers, n, l, ml for this electron. (will be marked) n = 5, l = 2, ml = -2 n = 5, l = 2, ml = -1 n = 5, l = 2, ml = 0 n = 5, l = 2, ml = 1 n = 5, l = 2, ml = 2 7. State which of the following orbitals cannot exist according to the quantum theory: 2s, 2d, 3p, 3f, 4f, and 5s. Briefly explain your answer. (will be marked) 2d and 3f cannot exist. The angular momentum quantum number l can only have integer numbers from 0 to n-l. For n= 2, l = 1 (p orbital) is the highest number for l. For n = 3, l = 2 (d orbital) is the highest number for l. 2