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From users who are members of CHY1201 - Spectroscopy :

Nick Walker

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6 years, 1 month ago

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Nick Walker 5 years, 11 months ago

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Nick Walker 6 years, 1 month ago

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Nick Walker 6 years, 1 month ago

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Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

n	number	2
Parameters	list	List of 4 items
Data	list	List of 3 items
randomiser	integer	0
Molecule	string	1,3-butadiene
Length	number	5.6
me	number	0
me_mantissa	number	9.109
me_log	integer	-31
h	number	0
h_mantissa	number	6.626
h_log	integer	-34
E_homo	number	0
E_homo_log	integer	-19
E_homo_mantissa	number	7.68
E_lumo	number	0
E_lumo_log	integer	-18
E_lumo_mantissa	number	1.73
Delta_E	number	0
Delta_E_log	integer	-19
Delta_E_mantissa	number	9.61
Frequency	number	1449675834370279.5
Frequency_Mantissa	number	1.45
Frequency_log	integer	15
Wavelength	number	207
ne	number	4
LUMO	number	3
Wavelength_m	number	0.000000207
Wavelength_m_log	integer	-7
wavelength_m_mantissa	number	2.07
test	number	0.0187969394

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
Data[2]

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `number`

← Depends on:

Data

→ Used by:

E_homo
E_lumo
LUMO
ne

Advice

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Part b) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Part c) Gap-fill
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Part d) Gap-fill
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Part e) Number entry

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

What is the energy, E (in J), of the HOMO of {Molecule}* if it is assumed that the "particle-in-a-box model" can be applied and the length of the box, l, is {Length} $\unicode{x212B}$ ?

Gap 0. $\times$ 10^{Gap 1.}

*If you are asked a question about beta-carotene, assume that there are 22 electrons in its conjugated system.

Advice

a) Each carbon atom contributes one electron to the conjugated system. Electrons are paired within molecular orbitals. Therefore, after adding the {ne} electrons of {Molecule} to molecular orbitals (filling the lowest energy orbitals first), we discover that the HOMO is {n}.

We need to remember that 1 $\unicode{x212B}$ in length is equal to 1 $\times$ 10^-10 m.

Next, we can note that;

E (n) = \frac{n^{2} h^{2}}{8 m_{e}^{2} L^{2}} = \frac{{n}^{2} \times ({h_mantissa} \times 10^{{h_log}} J s)^{2}}{8 \times {me_mantissa} \times 10^{{me_log}} k g \times ({Length} \times 10^{- 10})^{2} m^{2}}

$E(n)=\frac{n^2h^2}{8m_e^2L^2}=\frac{\var{n}^2\times (\var{h_mantissa}\times~10^{\var{h_log}}{\rm~J~s})^2}{8~\times\var{me_mantissa}~\times~10^{\var{me_log}}~{\rm kg}~\times~(\var{Length}~\times~10^{-10})^2~{\rm m^2}}$

It's a good idea to rearrange this to collect powers of ten;

\frac{{n}^{2} \times ({h_mantissa} \times 10^{{h_log}} J s)^{2}}{8 \times {me_mantissa} \times 10^{{me_log}} k g \times ({Length} \times 10^{- 10})^{2} m^{2}}

$\frac{\var{n}^2~\times~(\var{h_mantissa}\times~10^{\var{h_log}}{\rm~J~s})^2}{8~\times~\var{me_mantissa}\times~10^{\var{me_log}}~{\rm kg}~\times~(\var{Length}\times ~10^{-10})^2~{\rm m^2}}$

= \frac{{n}^{2} \times {h_mantissa}^{2}}{8 \times {me_mantissa} \times {Length}^{2}} \times \frac{(10^{{h_log}})^{2}}{10^{{me_log}} \times (10^{- 10})^{2}} J

$=\frac{\var{n}^2\times\var{h_mantissa}^2}{8\times\var{me_mantissa}\times\var{Length}^2}\times\frac{(10^{\var{h_log}})^2}{10^{\var{me_log}}\times(10^{-10})^2}~{\rm~~~ J}$

= {E_homo_mantissa} \times 10^{{E_homo_log}} J

$=\var{E_homo_mantissa}\times10^\var{E_homo_log}~{\rm~J}$

b) Electrons are paired within molecular orbitals. After adding the {ne} electrons of {Molecule} to molecular orbitals (filling the lowest energy orbitals first), we discover that the LUMO is {LUMO}.

We need to remember that 1 $\unicode{x212B}$ in length is equal to 1 $\times$ 10^-10 m.

Next, we can note that;

E (n) = \frac{(n + 1)^{2} h^{2}}{8 m_{e}^{2} L^{2}} = \frac{{LUMO}^{2} \times ({h_mantissa} \times 10^{{h_log}} J s)^{2}}{8 \times {me_mantissa} \times 10^{{me_log}} k g \times ({Length} \times 10^{- 10})^{2} m^{2}}

$E(n)=\frac{(n+1)^2h^2}{8m_e^2L^2}=\frac{\var{LUMO}^2~\times~(\var{h_mantissa}~\times~10^{\var{h_log}}{\rm~J~s})^2}{8~\times~\var{me_mantissa}~\times~ 10^{\var{me_log}}~{\rm kg}~\times~(\var{Length}~\times~10^{-10})^2~{\rm m^2}}$

It's a good idea to rearrange this to collect powers of ten;

\frac{{LUMO}^{2} \times ({h_mantissa} \times 10^{{h_log}} J s)^{2}}{8 \times {me_mantissa} \times 10^{{me_log}} k g \times ({Length} \times 10^{- 10})^{2} m}

$\frac{\var{LUMO}^2~\times~(\var{h_mantissa}~\times~10^{\var{h_log}}{\rm~J~s})^2}{8~\times~\var{me_mantissa}~\times~10^{\var{me_log}}~{\rm kg}~\times~ (\var{Length}~\times~10^{-10})^2~{\rm m}}$

= \frac{{LUMO}^{2} \times {h_mantissa}^{2}}{8 \times {me_mantissa} \times {Length}^{2}} \times \frac{(10^{{h_log}})^{2}}{10^{{me_log}} \times (10^{- 10})^{2}} J

$=\frac{\var{LUMO}^2\times\var{h_mantissa}^2}{8~\times~\var{me_mantissa}~\times~\var{Length}^2}~\times~\frac{(10^{\var{h_log}})^2}{10^{\var{me_log}}~\times~(10^{-10})^2}~{\rm~~~ J}$

= {E_lumo_mantissa} \times 10^{{E_lumo_log}} J

$=\var{E_lumo_mantissa}\times10^\var{E_lumo_log}~{\rm~J}$

c) Perhaps the most direct way to the answer to this question is simply to subtract the answer to (a) from that of (b);

{E_lumo_mantissa} \times 10^{{E_lumo_log}} J - {E_homo_mantissa} \times 10^{{E_homo_log}} J = {Delta_E_mantissa} \times 10^{{Delta_E_log}} J

$\var{E_lumo_mantissa}\times10^\var{E_lumo_log}~{\rm~J}-\var{E_homo_mantissa}\times10^\var{E_homo_log}~{\rm~J}=\var{Delta_E_mantissa}~\times~10^{\var{Delta_E_log}}~{\rm~J}$

but it is worth knowing that there is a more general route to a solution that avoids the need to calculate the energies of the HOMO and LUMO;

E (n + 1) - E (n) = \frac{(n + 1)^{2} h^{2}}{8 m_{e}^{2} L^{2}} - \frac{n^{2} h^{2}}{8 m_{e}^{2} L^{2}}

$E(n+1)-E(n)=\frac{(n+1)^2h^2}{8m_e^2L^2}-\frac{n^2h^2}{8m_e^2L^2}$

= (\frac{(n^{2} + 2 n + 1) h^{2}}{8 m_{e} L^{2}}) - (\frac{n^{2} h^{2}}{8 m_{e} L^{2}})

$=(\frac{(n^2+2n+1)h^2}{8m_eL^2})-(\frac{n^2h^2}{8m_eL^2})$

= \frac{(2 n + 1) h^{2}}{8 m_{e} L^{2}}

$=\frac{(2n+1)h^2}{8m_eL^2}$ so

\frac{(2 \times {n} + 1) \times {h_mantissa}^{2}}{8 \times {me_mantissa} \times {Length}^{2}} \times \frac{(10^{{h_log}})^{2}}{10^{{me_log}} \times (10^{- 10})^{2}} J

$\frac{(2~\times~\var{n}+1)~\times~\var{h_mantissa}^2}{8~\times~\var{me_mantissa}~\times~\var{Length}^2}~\times~\frac{(10^{\var{h_log}})^2}{10^{\var{me_log}}~\times~(10^{-10})^2}~{\rm~~~ J}$

= {Delta_E_mantissa} \times 10^{{Delta_E_log}} J

$=\var{Delta_E_mantissa}~\times~10^{\var{Delta_E_log}}~{\rm~J}$

d) We need to take the energy associated with the transition and divide it by the Planck constant;

E = h ν

$E=h\nu$

ν = \frac{E}{h} = \frac{{Delta_E_mantissa} \times 10^{{Delta_E_log}} J}{{h_mantissa} \times 10^{{h_log}} J s}

$\nu=\frac{E}{h}=\frac{\var{Delta_E_mantissa}~\times~10^{\var{Delta_E_log}}~{\rm J}}{\var{h_mantissa}~\times~10^{\var{h_log}}{\rm~J~s}}$

= {Frequency_mantissa} \times 10^{{Frequency_log}} H z

$={\var{Frequency_mantissa}\times10^{\var{Frequency_log}}~{\rm~Hz}}$

e) To obtain the wavelength from the frequency, we need to remember that;

c = λ ν

$c=\lambda\nu$

\frac{c}{ν} = λ = \frac{3.0 \times 10^{8} m s^{- 1}}{{Frequency_mantissa} \times 10^{{Frequency_log}} s^{- 1}} = {Wavelength_m_mantissa} \times 10^{{Wavelength_m_log}} m

$\frac{c}{\nu}=\lambda=\frac{3.0~\times~10^8~{\rm m~s^{-1}}}{\var{Frequency_mantissa}~\times~10^{\var{Frequency_log}}{\rm~s^{-1}}}=\var{Wavelength_m_mantissa}~\times~10^\var{Wavelength_m_log}{\rm~m}$

and noting that 1 $\times~10^9$ nm = 1 m, we can note that;

{Wavelength_m_mantissa} \times 10^{{Wavelength_m_log}} m = {Wavelength} n m

$\var{Wavelength_m_mantissa}~\times~10^\var{Wavelength_m_log}{\rm~m}=\var{Wavelength}{\rm~nm}$

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This question is used in the following exams:

Spectroscopy practice by Nick Walker in CHY1201 - Spectroscopy.
QUIZ #1 by Van-Oanh Nguyen-Thi in Van-Oanh's workspace.
"Particle in a Box" example questions by Nick Walker in CHY1201 - Spectroscopy.

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Nick Walker 5 years, 11 months ago

Nick Walker 6 years, 1 month ago

Nick Walker 6 years, 1 month ago

Error in variable testing condition

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Generated value: `number`

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Energy- PIAB

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Nick Walker 5 years, 11 months ago

Nick Walker 6 years, 1 month ago

Nick Walker 6 years, 1 month ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`