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In my textbook it said the following:

Photons with wavelengths in the spectral range of $[94\mathrm{\ nm},104\mathrm{\ nm}]$, interact the hydrogen atom in the basic state. Photons having those wavelengths can stimulate the hydrogen atom to $n=3,4,5$ levels.

I'm trying to figure out why it's true. Given some wavelength $\lambda$, how can I know which level it can get?

I'm familiar with the Rydberg formula: $$ \frac{1}{\lambda_{m\to n}}=R\cdot\left(\frac{1}{n^2}-\frac{1}{m^2}\right) $$ where $m>n$ and $R=1.097\cdot10^7\ \mathrm{m}^{-1}$. But because there are two values $n,m$, I'm struggling to figure out a sophisticated way to find the levels. I could just insert $\lambda=94\mathrm{\ nm}$ and check for each $n$ it's $m$'s but it sounds like not so much a sophisticated way. Is there a better way?

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Most atoms (speaking from a chemistry perspective, other situations may differ) are in or near the ground state. As such it is OK to simply consider (for example) $n = 1,2,3$, and solve those few cases for $m$. Observations will likely be dominated by the $n = 1$ case with a small contribution from the higher levels.

In this case $λ_{m \leftarrow 1} = \frac{1}{R(1-m^{-2})} = 102 \text{ nm}$ for $m = 3$, 97 nm for $m = 4$, etc, as you have seen.

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  • $\begingroup$ Wait, the Rydberg formula gives us the $\lambda_{m\to n}$ for $(m>n)$ or for $(m<n)$? $\endgroup$
    – vesii
    Oct 11, 2020 at 12:25
  • $\begingroup$ @vesii You want a positive result so with the formula as given in your question $m > n$. If the atom is being excited then it must be going to $m$: $λ_{m \leftarrow n}$. If the atom is instead emitting light the symbol would be $λ_{n \leftarrow m}$, and this would be numerically the same value. You can write the arrow pointing rightwards if your textbook/course prefers. $\endgroup$
    – GKFX
    Oct 11, 2020 at 12:32
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    $\begingroup$ So if we are going up the levels from $n$ to $m$, then $m>n$ and then $λ_{m \leftarrow n}=R\cdot\left(\frac{1}{n^2}-\frac{1}{m^2}\right)$. But when we are going down the levels, from $m$ to $n$, how the formula would look like for $λ_{n \leftarrow m}$? is it $λ_{n \leftarrow m}=R\cdot\left(\frac{1}{m^2}-\frac{1}{n^2}\right)$? or is it the same equation but different symbol in $\lambda$? In other words, $λ_{n \leftarrow m}=λ_{m \leftarrow n}$? $\endgroup$
    – vesii
    Oct 11, 2020 at 12:35
  • $\begingroup$ It's the same formula regardless of whether you are going up or down by the conservation of energy, giving the amount of energy lost going down or gained going up. The energy of a hydrogen atom (without considering relativity), with principle quantum number $n$, is $\frac{h c R}{n^2}$. The Rydberg formula is simply the difference between two copies of this formula, with the factor $hc$ removed to give inverse wavelength rather than energy. $\endgroup$
    – GKFX
    Oct 11, 2020 at 12:44
  • $\begingroup$ So the energy lost going down or gained going up is equal right? $\endgroup$
    – vesii
    Oct 11, 2020 at 12:45

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