Download Assignment 3 - SOLUTIONS

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.
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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.
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