(a) Interpretation: The wavelength of light corresponding to 1.00 × 10 − 32 J energy is to be stated. The comparison of this wavelength to the diameter of the Earth, which is 1.27 × 10 7 m , is to be shown. Concept introduction: The energy for particle in a box is given by the expression as follows. E = n 2 h 2 8 m a 2 Where, • E is the energy of the particle • n is the number of energy level • h is Planck’s constant • m is the mass of the particle • a is the width of the box The expression of the energy for particle in a box involves n which shows that the energy is quantized for a particle in a box.
(a) Interpretation: The wavelength of light corresponding to 1.00 × 10 − 32 J energy is to be stated. The comparison of this wavelength to the diameter of the Earth, which is 1.27 × 10 7 m , is to be shown. Concept introduction: The energy for particle in a box is given by the expression as follows. E = n 2 h 2 8 m a 2 Where, • E is the energy of the particle • n is the number of energy level • h is Planck’s constant • m is the mass of the particle • a is the width of the box The expression of the energy for particle in a box involves n which shows that the energy is quantized for a particle in a box.
Solution Summary: The author explains that the energy for particle in a box is given by the expression as follows.
The wavelength of light corresponding to 1.00×10−32J energy is to be stated. The comparison of this wavelength to the diameter of the Earth, which is 1.27×107m, is to be shown.
Concept introduction:
The energy for particle in a box is given by the expression as follows.
E=n2h28ma2
Where,
• E is the energy of the particle
• n is the number of energy level
• h is Planck’s constant
• m is the mass of the particle
• a is the width of the box
The expression of the energy for particle in a box involves n which shows that the energy is quantized for a particle in a box.
Interpretation Introduction
(b)
Interpretation:
The width of a box that an electron needs to be in, to possess 1.00×10−32J energy is to be calculated.
Concept introduction:
The energy for particle in a box is given by the expression as follows.
E=n2h28ma2
Where,
• E is the energy of the particle
• n is the number of energy levels
• h is Planck’s constant
• m is the mass of the particle
• a is the width of the box
The expression of the energy for particle in a box involves n which shows that the energy is quantized for a particle in a box.
(f) SO:
Best Lewis Structure
3
e group geometry:_
shape/molecular geometry:,
(g) CF2CF2
Best Lewis Structure
polarity:
e group arrangement:_
shape/molecular geometry:
(h) (NH4)2SO4
Best Lewis Structure
polarity:
e group arrangement:
shape/molecular geometry:
polarity:
Sketch (with angles):
Sketch (with angles):
Sketch (with angles):
1.
Problem Set 3b
Chem 141
For each of the following compounds draw the BEST Lewis Structure then sketch the molecule (showing
bond angles). Identify (i) electron group geometry (ii) shape around EACH central atom (iii) whether the
molecule is polar or non-polar (iv)
(a) SeF4
Best Lewis Structure
e group arrangement:_
shape/molecular geometry:
polarity:
(b) AsOBr3
Best Lewis Structure
e group arrangement:_
shape/molecular geometry:
polarity:
Sketch (with angles):
Sketch (with angles):
(c) SOCI
Best Lewis Structure
2
e group arrangement:
shape/molecular geometry:_
(d) PCls
Best Lewis Structure
polarity:
e group geometry:_
shape/molecular geometry:_
(e) Ba(BrO2):
Best Lewis Structure
polarity:
e group arrangement:
shape/molecular geometry:
polarity:
Sketch (with angles):
Sketch (with angles):
Sketch (with angles):
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