Using dimensional analysis to ‘get’ planck’s ${\lambda}$ equation with avogadro’s constant

Question

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 ${\lambda}$

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

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

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

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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?

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Units:

${\lambda=m}$

${E=j}$

${h=js}$

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

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

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Dimensional analysis:

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

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

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Then I can’t get ${\lambda=m}$

Is this correct, and if not then why?:

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

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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

Answer 1

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:

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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:

$$\text [m]$$

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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:

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

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

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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.

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Units:

${\lambda=\text m}$

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

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

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

$N_A=mol^{-1}$

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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:

$$ [\lambda=m] $$

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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:

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

Answer 2

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.

Answer 3

$${E=j\cdot \frac{1}{n}\Rightarrow{E=j}}$$

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.

Answer 4

$\require{cancel}$

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 $N_A$

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