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This is for when you calculate bond energy in: ${j\cdot mol(n)^{-1}}$ and want to use planck’s equation for photon energy to calculate ${\lambda}$


Planck’s equation for photon energy: $${\lambda=\frac{h\cdot c}{E}}$$

Planck’s equation for photon energy with avogadro’s constant:

$${\lambda=\frac{h\cdot c\cdot N_A}{E}}$$


I don’t understand how ${N_A}$ (${\frac{1}{n}}$) is at the top as:

${n=\frac{g}{g\cdot n^{-1}}\Rightarrow{n=\frac{1}{n}}}$

Energy in this case:

${E=\frac{j}{mol\ or\ n \ (1/n)}}$

I can’t rearrange ${1/n}$ to get it to the to the top in order to get rid of the ${n}$ at the bottom so that:

${E=j}$

Can someone show me verbosely with dimensional analysis how to do this?


Units:

${\lambda=m}$

${E=j}$

${h=js}$

${c=ms^{-1}}$

${N_A=n^{-1}}$


Dimensional analysis:

${\lambda=js\cdot ms^{-1}\cdot j^{-1}(1^{-1}n^{-2})}$

${\lambda=\frac{j\cdot s\cdot m}{j\cdot s(\frac{1}{n})}}\Rightarrow{\lambda=\frac{j\cdot s\cdot m}{\dfrac {j\cdot s}{n}}}\Rightarrow{\lambda=\frac{m}{\dfrac{1}{n}}}$


Then I can’t get ${\lambda=m}$

Is this correct, and if not then why?:

${\frac{j}{mol(1/n)}=\frac{j}{\dfrac{1}{n}}}$


Bonus question:

I have been using dimensional analysis to ‘get’ equations before I even knew that word, as my memory is bad and I couldn’t remember equations, I used to think it was a trick

Are there more/better ways to ‘get’ equations and cases where dimensional analysis won’t work?

I really need help here and would appreciate it very much

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    $\begingroup$ You cannot just "insert" a new variable into an equation from physics/chemistry just to get a desired result. You should have some reason for this. And the reason is that in the equation: $$\lambda=\frac{hc}{E}$$ the energy is given per bond ($\approx$ per molecule). There is no hope to obtain the correct result if you insert instead the energy per mole. Now try to guess how to calculate the energy per molecule from the energy per mole (hint: use different symbols or indices for these energies). $\endgroup$
    – user
    Mar 15, 2021 at 12:06
  • $\begingroup$ Please look at my edit to see my logic and what I’m ttying to do @user $\endgroup$
    – Nickotine
    Mar 15, 2021 at 12:31
  • $\begingroup$ Please read carefully my previous comment ant try to find a relationship between the energy per molecule ($E_m$) and the energy per mole ($E_M$). $\endgroup$
    – user
    Mar 15, 2021 at 12:33

4 Answers 4

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Maybe I wrongly understand the question but: $$ [\text m]=[\lambda]\stackrel{?}=\left[\frac{hc N_A}{E}\right]=\frac{[\text J\cdot\text s][\text m\cdot \text s^{-1}][\text{mol}^{-1}]}{[\text J\cdot\text{mol}^{-1}]}=[\text m]. $$ so that there is no inconsistency in this equation.

And of course the inconsistency will appear if $N_A$ is absent or wrongly placed.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Mar 16, 2021 at 22:09
  • $\begingroup$ Thanks for the answer but I didn’t ask for this at all, any guy on the street can do this, the only good thing I got from you is that $$\text n\ !=mol$$ so thanks for that, but otherwise this answer completely missing the point @user $\endgroup$
    – Nickotine
    Mar 17, 2021 at 23:39
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$\require{cancel}$

Let's assume for the following identity to hold:

$$\star:\left\{\lambda=\frac{hc N_A}{E}\right\}$$

Which expressed in units of measurement is:

$$\lambda=\frac{hc N_A}{E}=\frac{[\text J\cdot\text s][\text m\cdot \text s^{-1}][\text{mol}^{-1}]}{[\text J\cdot\text{mol}^{-1}]}=[\text m]$$

where $$E=[\text{J}\cdot\text{mol}^{-1}]$$

Where the 'naked' variables are quantities, and the square bracket-ed variables are units of measurements.

Let's divide both sides of the equation$(\star)$ by $\color{blue}{N_A}$ as per the rules of algebra.

$$\frac{\lambda}{\color{blue}{N_A}}=\frac{hc {N_A}}{E{\color{blue}{N_A}}}$$

Then we can simplify by $N_A$

$$\frac{\lambda}{\color{blue}{N_A}}=\frac{hc \cancel{N_A}}{E\cancel{\color{blue}{N_A}}} \Leftrightarrow \frac{\overbrace{[\text J\cdot\text s]}^h \overbrace{[\text m\cdot \text s^{-1}]}^c \overbrace{\cancel{[\text{mol}^{-1}]}}^{N_A}}{\underbrace{[\text J\cdot\text{mol}^{-1}]}_{E} \underbrace{\cancel{[\text{mol}^{-1}]}}_{\color{blue}{N_A}}}=\frac{[\text m]}{[\text{mol}^{-1}]}$$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Mar 16, 2021 at 22:09
  • $\begingroup$ This makes perfect sense, without [$N_A$] you get $$\lambda=\frac{m}{mol^{-1}}$$ hence why need [$N_A$] at the top so $$\lambda=m$$ @LordCommander $\endgroup$
    – Nickotine
    Mar 17, 2021 at 23:22
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    $\begingroup$ Okay, so what's our confusion again? 😂 $\endgroup$ Mar 17, 2021 at 23:23
  • $\begingroup$ Look at my latest answer if you still don’t believe me that they’re equivalent or at: $$ \lambda=\frac{\text h\cdot \text c}{j}=\frac{\text h\cdot \text c\cdot \frac{1}{mol}}{\text j\cdot \frac{1}{mol}}=\frac{\text h\cdot \text c\cdot 1}{\text j\cdot 1} $$ $\endgroup$
    – Nickotine
    Mar 17, 2021 at 23:24
  • $\begingroup$ Looks like now we agree 🙃 $\endgroup$
    – Nickotine
    Mar 17, 2021 at 23:27
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$${E=j\cdot \frac{1}{n}\Rightarrow{E=j}}$$

This does not ever happen. In the same framework of $E$ this is impossible (unless of course if $1/n$ equals $1$)

The two lambdas described in

${\lambda=\frac{h\cdot c}{E}}$

and

${\lambda=\frac{h\cdot c\cdot N_A}{E}}$

are NOT the same equation. Using the same variable($\lambda$) for two different quantities without context is negligent.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Mar 16, 2021 at 22:09
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I now know where I should have inserted [$N_A$] using dimensional analysis.

When I asked the below question I though that $\text [n]$ is just another way of saying ${[mol]}$.

My logic was sound, let me just replace $\text [n]$ with $[mol]$ this is how the original question should have been asked:


This is for when you calculate bond energy in:

$$ \text [j]\cdot [mol^{-1}] $$

Then use planck’s equation for photon energy to calculate [${\lambda}$] in:

$$\text [m]$$


Planck’s equation for photon energy:

$$ \lambda=\frac{\text h\cdot \text c}{\text E} $$

Planck’s equation for photon energy with avogadro’s constant:

$$ E=\text j\cdot \frac{1}{mol} $$

$$ \lambda=\frac{\text h\cdot \text c\cdot N_A}{\text E} $$


I now know how I could’ve known where to insert [$N_A$] using dimensional analysis only, just how I wanted.

This does not require any physics or chemistry knowledge aside from [$N_A$] which all you mathmeticians seemed to know about.

So it’s a maths question, onto the dimensional analysis part in my original question.


Units:

${\lambda=\text m}$

$\text E=\text j\cdot mol^{-1}$

$\text h= \text j\cdot \text s$

$\text c= \text m\cdot \text s^{-1}$

$N_A=mol^{-1}$


Dimensional analysis:

$$ \lambda=\frac{\text j\cdot \text s\cdot \text m}{\text j\cdot \text s(\frac{1}{mol})}\Rightarrow\lambda=\frac{\text j\cdot \text s\cdot \text m}{\dfrac {\text j\cdot \text s}{mol}}\Rightarrow\lambda=\frac{\text m}{\dfrac{1}{mol}} $$

Then I can’t get:

$$ [\lambda=m] $$


This was the dimensional analysis of planck’s [$\lambda$] equation without [$N_A$]:

$$ E=j\cdot \frac{1}{mol} $$

$$ \lambda=\frac{\text h\cdot \text c}{\text E} $$

So of course the dimensional analysis gave me:

$$ \lambda=\frac{\text m}{\dfrac{1}{mol}} $$

But it told me I need to put [$N_A$] at the top which I should have seen.

The whole reason for me wanting to add [$N_A$] was to get rid of [$mol$] from [$j$] in the [$E$] term, I just didn’t know where to put it, but it was made clear...

Now if I had noticed this and put [$N_A$] at the top then new dimensional analysis:

$$ \lambda=\frac{\text j\cdot \text s\cdot \text m\cdot (\frac{1}{mol})}{\text j\cdot \text s\cdot (\frac{1}{mol})}\Rightarrow\lambda=\frac{\text j\cdot \text s\cdot \text m}{\dfrac {\text j\cdot \text s\cdot mol}{mol}}\Rightarrow\lambda=\frac{\text m}{\dfrac{mol}{mol}}\Rightarrow\lambda=\text m $$

So:

$$ \lambda=\frac{\text h\cdot \text c}{\text E}=\frac{\text h\cdot \text c\cdot N_A}{\text E} $$

As:

$$ \lambda=\frac{\text h\cdot \text c}{j}=\frac{\text h\cdot \text c\cdot \frac{1}{mol}}{\text j\cdot \frac{1}{mol}}=\frac{\text h\cdot \text c\cdot 1}{\text j\cdot 1} $$

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  • $\begingroup$ I see, in this sense the $[N_A]$-less equation was wrong. $\endgroup$ Mar 17, 2021 at 23:30
  • $\begingroup$ Yes of course, which is why I was trying to add $N_A$ in because like you said having $$E=j\cdot \frac{1}{mol}$$ is crazy, I just didn’t know where to add it, now I get it completely, btw apart from dimensional analysis are there other ways to ‘make up equations? And your mathjaxx is on a whole other level, this stuff seriously needs an editor, do you just do all that stuff manually on here? @LordCommander $\endgroup$
    – Nickotine
    Mar 17, 2021 at 23:34
  • $\begingroup$ :), Make up equations? It's just practise. I have started doing latex documents in september. You just learn stuff on the go, \color{red}{stuff}, \cancel{stuff to cancel} and you have to include '\require{cancel}' at math.stackexchange, \underbar{stuff}_{the thing under} $\endgroup$ Mar 17, 2021 at 23:40
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    $\begingroup$ Of course all of this in between $ signs. Editing a latex document is not so much harder. But you may be able to packages that are not included here on math.stackexchange. It's also offline, yours and much more customizable(obviously). If you wanna understand what code i use just right click on my equation and choose MathJax or view my answers in edit mode. $\endgroup$ Mar 17, 2021 at 23:51
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    $\begingroup$ Well that’s exactly how I got some stuff from you :) $\endgroup$
    – Nickotine
    Mar 17, 2021 at 23:53

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