Estimate the following quantities from a graph of the refractive index of NaCl

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SUMMARY

The discussion focuses on estimating quantities related to the refractive index of sodium chloride (NaCl) using its dielectric properties. Key values derived include the static permittivity, εst = 5.78 and ε = 1.69, with the angular frequency ω0 determined as 5 × 1012 Hz. The participants also explore the relationship between force, spring constant, and angular frequency, leading to confusion regarding the calculation of reduced mass, which is critical for accurate results. The conversation highlights the importance of distinguishing between angular frequency and frequency in calculations.

PREREQUISITES
  • Understanding of dielectric constants, specifically εst and ε.
  • Knowledge of angular frequency (ω) and its relation to frequency (f).
  • Familiarity with Hooke's law and its application in ionic solids.
  • Ability to calculate reduced mass using the formula μ = m1m2 / (m1 + m2).
NEXT STEPS
  • Research the derivation of the relationship between εst, ε, and the number density of molecules in ionic solids.
  • Learn about the differences between angular frequency and frequency in physical systems.
  • Explore the concept of reduced mass in detail, particularly in the context of diatomic molecules.
  • Investigate the application of Hooke's law in estimating forces in ionic crystals.
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Students and researchers in solid-state physics, materials science, and anyone involved in the study of ionic compounds and their dielectric properties.

mcas
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Homework Statement
The attached plot represents a refractive index for NaCl. Resonance is caused by oscillations of dipoles Na+Cl-. Atomic numbers for Na and Cl are equal to respectively 23 and 35.5.
Estimate:
a) [itex]\varepsilon_{st}[/itex] and [itex]\varepsilon_{\infty}[/itex]
b) resonance frequency [itex]\omega_0[/itex]
c) restoring force for a unit displacement
d) concentration of NaCL molecules.

Answers: a) 5.9, 1.96 b) [itex]5\cdot 10^{12} Hz[/itex], c) 23 N, d) [itex]3\cdot 10^{28} m^{-3}[/itex]
Relevant Equations
1. [itex]n=\frac{1}{\sqrt{2}} (\varepsilon_{1} + \sqrt{\varepsilon_{1}^2 + \varepsilon_{2}^2})^{1/2}[/itex]
where [itex]\varepsilon_{1}[/itex] is a real part of a complex dielectric function and [itex]\varepsilon_{2}[/itex] is an imaginary part.

2. [itex]k=m\omega_0^2[/itex]
plot.png


a) I managed to obtain some results that are roughly around what is given in the answers.
Because \varepsilon_{st} and \varepsilon_{\infty} are values of \varepsilon_{1}, I used this approximation:
n\approx \frac{1}{\sqrt{2}} (\varepsilon_{1}+\sqrt{\varepsilon_{1}^2})^{1/2}
-> \varepsilon_{1} = n^2
And that gave me:
\varepsilon_{st} = 5.78
\varepsilon_{\infty}=1.69

I'm not sure if this is the right approach.

b) That's an easy one, I hot that right from the plot:
\omega_0 = 5 \cdot 10^{12} \ Hz

c) As F=kx, I used the relation k=m\omega_0^2.
However, this gave me a wrong result of 2.42 N. I'm not sure where I made an error.

d) With this one, I don't even know where to start.

I would really appreciate your help. This is a very tricky problem for me.
 
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mcas said:
a) I managed to obtain some results that are roughly around what is given in the answers.
Because \varepsilon_{st} and \varepsilon_{\infty} are values of \varepsilon_{1}, I used this approximation:\varepsilon_{1} = n^2And that gave me:
\varepsilon_{st} = 5.78
\varepsilon_{\infty}=1.69
It will help if you tell us what values you used for ##n_{st}## and ##n_{\infty}## and how you got those values. Also, it's helpful to explain the meaning of the subscripts st, ##\infty##, and ##1##.

mcas said:
b) That's an easy one, I got that right from the plot:
\omega_0 = 5 \cdot 10^{12} \ Hz
The notation ##\omega## usually denotes angular frequency (in rad/s) rather than frequency ##f## (in Hz).

mcas said:
c) As F=kx, I used the relation k=m\omega_0^2.
However, this gave me a wrong result of 2.42 N. I'm not sure where I made an error.
Here, ##\omega## is angular frequency. What value did you use for the mass ##m##? Show how you got this value.

mcas said:
d) With this one, I don't even know where to start.

I would really appreciate your help. This is a very tricky problem for me.
There might be different ways to approach this part. They only want 1 significant figure, so you can try some sort of estimation. If you displace one of the ions by a distance ##x## toward one of its nearest neighbors, find a rough approximation for the restoring force on the ion due to electrical forces. Can you express the restoring force as being proportional to ##x## so that you have Hooke's law?

EDIT: Oops, I wasn't careful here. If you consider only the two nearest neighbors along the line of the displacement, then the force is not restoring. The force is away from the equilibrium point! To get a restoring force will require taking into account interactions with more distant ions rather than just the nearest neighbors. But, that might get messy. Maybe someone else can offer suggestions here.
 
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Regarding part (d):

Have you covered any formulas that relate ##\varepsilon_{st}, \varepsilon_{\infty}##, and the number density of molecules for ionic solids? I have a solid-state textbook that derives such a formula. Using data for this problem, I was able to get a value for the number density of molecules that agrees with the stated answer.
 
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TSny said:
It will help if you tell us what values you used for ##n_{st}## and ##n_{\infty}## and how you got those values. Also, it's helpful to explain the meaning of the subscripts st, ##\infty##, and ##1##.
I got these values from the attached figure: n_{st}=2.4 and n_{\infty}=1.3.
Ok, now the explaining.
I define a complex dielectric function (permittivity) as
\widetilde{\varepsilon} = \varepsilon_1 + i \varepsilon_2
So \varepsilon_1 is the real part of the dielectric function and \varepsilon_2 is the imaginary part.
The subscript st comes from 'static' and it basically means that it's a value of the dielectric function for small frequencies. Analogically, \infty means that a subscripted value is given for big frequencies.
I'm also attaching a few figures to visualize this:
00.png


TSny said:
The notation ##\omega## usually denotes angular frequency (in rad/s) rather than frequency ##f## (in Hz).Here, ##\omega## is angular frequency. What value did you use for the mass ##m##? Show how you got this value.
Thank you for the comment about the notation, it's always good to be super diligent!
I've calculated the mass using the atomic numbers given in the homework statement.
(23+35.5) \cdot 1.66 \cdot 10^{-27} \ kg \approx 2.42 kg

TSny said:
Regarding part (d):

Have you covered any formulas that relate ##\varepsilon_{st}, \varepsilon_{\infty}##, and the number density of molecules for ionic solids? I have a solid-state textbook that derives such a formula. Using data for this problem, I was able to get a value for the number density of molecules that agrees with the stated answer.

I know this one:
\varepsilon_{st} - \varepsilon_{\infty} = \frac{Ne^2}{\varepsilon_0 m \omega_0^2}
but I believe N is a number of atoms, so I still don't know how to calculate the concentration.
I would be very thankful if you shared the formula you used.
 
mcas said:
I got these values from the attached figure: n_{st}=2.4 and n_{\infty}=1.3.

OK. I think they might have used 1.4 instead of 1.3 for ##n_{\infty}##. It's hard to be precise by just reading the graph.

mcas said:
Thank you for the comment about the notation, it's always good to be super diligent!
I've calculated the mass using the atomic numbers given in the homework statement.
(23+35.5) \cdot 1.66 \cdot 10^{-27} \ kg \approx 2.42 kg

In the relation ##k = m\omega_0^2##, ##m## represents the reduced mass of the two ions. The reduced mass should be smaller than either of the ion masses.
mcas said:
I know this one:
\varepsilon_{st} - \varepsilon_{\infty} = \frac{Ne^2}{\varepsilon_0 m \omega_0^2}
but I believe N is a number of atoms, so I still don't know how to calculate the concentration.
I would be very thankful if you shared the formula you used.

Ok, that looks the same as what was in my book. ##N## is the number density of the "molecules" and ##m## is the reduced mass.
 
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TSny said:
In the relation ##k = m\omega_0^2##, ##m## represents the reduced mass of the two ions. The reduced mass should be smaller than either of the ion masses.

Ok, that looks the same as what was in my book. ##N## is the number density of the "molecules" and ##m## is the reduced mass.
The last one got me confused because the lecturer just waved their hands saying "well you know it's the atoms". But that explains a lot.

Thank you! Really, I needed some guidance 😅 Have a nice day!
 
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TSny said:
Regarding part (d):

Have you covered any formulas that relate ##\varepsilon_{st}, \varepsilon_{\infty}##, and the number density of molecules for ionic solids? I have a solid-state textbook that derives such a formula. Using data for this problem, I was able to get a value for the number density of molecules that agrees with the stated answer.
Ok, I plugged in the numbers and I might still need a little help.
Using the result from the answers in c), I came to a conclusion that the reduced mass (I'm going to denote it by \mu must be equal to \mu = 9.2 \cdot 10^{-25} \ kg because F= kx = k\cdot 1 = \mu \omega_0^2 so we get 23 = 25 \cdot 10^{-24} \cdot \mu and that gives \mu = 9.2 \cdot 10^{-25} \ kg.
Which also gives a correct answer for d) when plugging it into the formula.

But when I want to calculate the reduced mass, I get a different result:
\mu = \frac{m_1m_2}{m_1+m_2}= \frac{23\cdot 35.5}{23+35.5}\approx 14 [ u ] = 14 \cdot 1.661 \cdot 10^{-27} \ [kg] \approx 2.35 \cdot 10^{-26} \ [kg]

I searched around the Internet and this is in fact the reduced mass of NaCl. So I think I still must get something wrong if you managed to get the result in d) the same as the answer.
 
mcas said:
Using the result from the answers in c), I came to a conclusion that the reduced mass (I'm going to denote it by \mu must be equal to \mu = 9.2 \cdot 10^{-25} \ kg because F= kx = k\cdot 1 = \mu \omega_0^2 so we get 23 = 25 \cdot 10^{-24} \cdot \mu and that gives \mu = 9.2 \cdot 10^{-25} \ kg.

Keep in mind that ##\omega_0## is an angular frequency. The horizontal axis in the graph is frequency ##f##, not ##\omega##.

mcas said:
But when I want to calculate the reduced mass, I get a different result:
\mu = \frac{m_1m_2}{m_1+m_2}= \frac{23\cdot 35.5}{23+35.5}\approx 14 [ u ] = 14 \cdot 1.661 \cdot 10^{-27} \ [kg] \approx 2.35 \cdot 10^{-26} \ [kg]
Ok. (I get a little bit smaller value of ## 2.32 \cdot 10^{-26}## kg if I wait until the end to round off to 3 sig figs. )
 
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The answer in the book for part (b) is not the correct answer for the angular frequency ##\omega_0##. The answer in the book would be the correct answer for the frequency ##f_0##.
 
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TSny said:
The answer in the book for part (b) is not the correct answer for the angular frequency ##\omega_0##. The answer in the book would be the correct answer for the frequency ##f_0##.
That's it! Thank you so much once again.
 
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