For planck’s photon energy equation when calculating wavelength it makes no sense for it to be negative.

The answer I get is negative because the energy value is the only negative variable, heat is given out so it’s said that the reaction is exothermic heat taken in is positive.


$h=6.262\cdot 10^{-34}\ j\cdot s^{-1}$

$c=2.998\cdot 10^{8}\ m\cdot s^{—1}$

${NA}=6.022\cdot 10^{23}\ mol^{-1}$

$E=-533000\ j\cdot mol^{-1}$

$\lambda=x\cdot m$

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

$$ \lambda=\frac{(6.626\cdot 10^{-34})(2.998\cdot 10^{8})(6.022\cdot10^{23})}{(-533000)}=-224nm $$

Obviously the length can’t be negative so I just so I just put it as positive, but is there a mathematical technique for situations like this? I’m just trying to learn.

  • 5
    $\begingroup$ All of those numbers are positive. I suspect your error is in using our calculator. $\endgroup$
    – John Douma
    Mar 27, 2021 at 16:05
  • $\begingroup$ damn the denominator was meant to be negative @johndouma I edited $\endgroup$
    – Nickotine
    Mar 27, 2021 at 16:10
  • $\begingroup$ @Nickotine How come the value of $\;j\cdot mol\;$ is negative ? $\endgroup$
    – DonAntonio
    Mar 27, 2021 at 16:12
  • $\begingroup$ This isn't really a math question (certainly not a linear-algebra question). Maybe the Physics StackExchange would be a better place for it. $\endgroup$
    – Blue
    Mar 27, 2021 at 16:12
  • $\begingroup$ it’s just how do I mathematically deal with this wrong negative value? @blue not sure what to tag this as $\endgroup$
    – Nickotine
    Mar 27, 2021 at 16:12

1 Answer 1


Just make one of the other values negative as well, for example, let $c=-2.998\cdot 10^{8}\ m\cdot s^{—1}$. Then you get a positive length.

  • $\begingroup$ but are you just ‘allowed’ to do that? @JMP $\endgroup$
    – Nickotine
    Mar 27, 2021 at 16:52
  • 1
    $\begingroup$ Well, $c=\pm\sqrt{\frac{E}{m}}$. so yes you are! $\endgroup$
    – JMP
    Mar 27, 2021 at 16:58
  • $\begingroup$ what the hell how did you come up with that? See I knew mathematicians would give me some crazy good answer that’s why I posted this here, thank you so much @JMP $\endgroup$
    – Nickotine
    Mar 27, 2021 at 17:03
  • $\begingroup$ oh just rearranging $$E=m\cdot c{2}$$ Very smart @JMP $\endgroup$
    – Nickotine
    Mar 27, 2021 at 22:51

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