<p>I don't quite understand your plot (does &quot;grey&quot; mean &quot;blackbody&quot;?), but I think you've stated what's going on pretty clearly (I'm reading <em>Fundamental Astronomy</em> by Karttumen et al to make sure).</p>
<p>The Balmer series starts with n=2 to n=3 at 656 nm, and as the final energy level increases, the lines get closer together (<a href="https://www.vcalc.com/wiki/ekskekel/Balmer-Rydberg+Equation" rel="nofollow noreferrer">I found this nifty calculator</a>, which gives answers that look right).</p>
<p><span class="math-container">$$n_f=5\qquad \lambda=434~nm$$</span>
<span class="math-container">$$n_f=6\qquad \lambda=410~nm$$</span>
<span class="math-container">$$n_f=7\qquad \lambda=397~nm$$</span>
<span class="math-container">$$...$$</span>
<span class="math-container">$$n_f=50,\qquad \lambda=365.091~nm$$</span>
<span class="math-container">$$n_f=51,\qquad \lambda =365.068~nm$$</span></p>
<p>And so forth, until the free state at 364.7 nm. Visually, this looks like some kind of continuum absorption, since the lines are not resolvable. But after that, you don't hit the next absorption line (Lyman) until 121.6 nm (that's n=1 to n=2). So calling that a jump (or discontinuity) is really more about the distribution of the lines - they go from being very close together to being regularly spaced apart again. Presumably we could also define &quot;Lyman jump&quot; at 91.2 nm, and Pachsen jump at 821 nm, etc.</p>

CC BY-SA 4.0
2023-02-01T20:21:48.260Z
This isn't really correct. The "jump" is an *increase* in absorption at wavelengths *shorter* than 364.7 nm, due to photoionization of $n = 2$ atoms. The crowding together of individual absorption lines with higher and higher values of $n_{f}$ modifies the shape of the jump immediately longward of 364.7 nm, but that's all.
Peter Erwin
CC BY-SA 4.0
2023-02-01T22:06:33.777Z
@PeterErwin: Ah yeah thanks, that's closer to what the OP was asking, that explains his plots. The description in Karttuman et al is more about the crowding of the lines, but yeah, once you *can* ionize from the n=2 state, you'll get excess absorption because of it.
cduston
CC BY-SA 4.0
2023-02-01T18:33:48.497Z
2023-02-01T18:33:48.497Z
cduston
<p>The Balmer break comes from a combination of two main things: the ability of photons with high enough energies (wavelengths shorter than 364.5 nm) to ionize hydrogen atoms that are in the <span class="math-container">$n = 2$</span> energy level, and the <em>inability</em> of any photons with lower energies (wavelengths longer than 364.5 nm) to ionize the atoms. This leads to an abrupt change in absorption of outgoing photons, from high absorption at wavelengths shorter than 364.5 nm, to no absorption at longer wavelengths.</p>
<p>Another important factor: Any photon with a wavelength shorter than 364.5 nm can ionize an <span class="math-container">$n = 2$</span> hydrogen atom -- but the probability (due to the cross-section of the photon-atom interaction) goes <em>down</em> as the wavelength gets shorter. That's why you get the sawtooth pattern: very-short-wavelength photons <em>can</em> ionize the atoms, but are unlikely to. As the wavelengths get longer, the probability of ionizing a photon increases, up to the point of 364.5 nm. Once the wavelength is longer than that, ionization is simply impossible, so the absorption goes to zero.</p>
<p>In practice, the <em>total</em> absorption doesn't go to zero, because there are <em>other</em> sources of absorption, including hydrogen atoms in the higher states, like the <span class="math-container">$n = 3$</span> state (which has its own &quot;Paschen jump&quot;), and things like H<span class="math-container">$^{-}$</span> hydrogen ions. You also have to think about things like how many atoms are in different energy levels. In a very hot atmosphere, most of the hydrogen atoms will be in higher energy levels (or even ionized), so you won't get much of a Balmer break.</p>
<p><em>do bound-bound absorptions not occur at all for stars with these discontinuities?</em></p>
<p>They absolutely do occur -- they produce <a href="https://en.wikipedia.org/wiki/Spectral_line" rel="nofollow noreferrer"><em>absorption lines</em></a>, such as those of the <a href="https://en.wikipedia.org/wiki/Balmer_series" rel="nofollow noreferrer">Balmer series</a> (which are not shown in the simplified, cartoon spectra in your figure). If you have enough hydrogen atoms in the <span class="math-container">$n = 2$</span> state to produce a noticeable Balmer break, then you'll automatically have Balmer absorption lines (H<span class="math-container">$\alpha$</span>, H<span class="math-container">$\beta$</span>, etc.) due to longer-wavelength photons being absorbed by the same atoms.</p>
<p>These lines do get closer together as you approach 364.5 nm from the long-wavelength side, which has the effect of making the peak of the sawtooth a bit rounded, but they don't produce the jump -- they make it a little bit more gradual. (Your figures do not show the effects of the absorption lines.)</p>

CC BY-SA 4.0
2023-02-03T05:31:24.773Z
Wow, this makes so much sense to me. Thank you very much for this detailed explanation.
Hika
CC BY-SA 4.0
2023-02-01T20:18:09.597Z
2023-02-01T20:18:09.597Z
Peter Erwin
<p>To quote <a href="https://en.wikipedia.org/wiki/Balmer_jump#:%7E:text=It%20is%20caused%20by%20electrons,wavelengths%20shorter%20than%20364.5%20nm." rel="noreferrer">Wikipedia</a>:</p>
<blockquote>
<p>Balmer Jump is caused by electrons being completely ionized directly from the second energy level of a hydrogen atom (bound-free absorption), which creates a continuum absorption at wavelengths shorter than 364.5 nm.</p>
</blockquote>
<p>Based on the graphs below we used in class, the non-grey energy distribution of the energy flux is continuously lower than the grey case for wavelengths shorter than <span class="math-container">$364.5nm$</span>, which according to Wikipedia is due to the bound-free absorptions of electrons in the <span class="math-container">$n=2$</span> energy state.</p>
<p>I understand why the energy flux gets lower for those wavelengths but I still cannot grasp why the discontinuity occurs. Is it because for wavelengths higher than <span class="math-container">$365.4nm$</span>, the bound-free absorption of electrons of <span class="math-container">$n=3$</span> starts? And do bound-bound absorptions not occur at all for stars with these discontinuities?</p>
<p>I would be very thankful for any clarification on this subject.</p>
<p><a href="https://i.sstatic.net/EWTtC.png" rel="noreferrer"><img src="https://i.sstatic.net/EWTtC.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-01-30T20:46:08.733Z
I wonder if there is any relationship to [Why does titanium oxide around Betelgeuse produce this particular sawtooth-shaped absorption spectrum?](https://astronomy.stackexchange.com/q/35451/7982)
uhoh
CC BY-SA 4.0
2023-02-01T16:43:48.070Z
Thank you for your comment.

Indeed the two spectra do have similar shapes, but reading the comments of the post mentioned it looks like the Betelgeuse's spectrum is due to the TiO behavior as a heavy metal, whereas the Balmer Jump is supposedly due to the bound-free transitions of n=2 Hydrogen electrons.

Thank you for the reference though, I will try to dig deeper that way as well.
Hika
<star><spectra><stellar-atmospheres>
What causes the Balmer Jump?
Hika
# What causes the Balmer Jump?

(Asked by: username_0 on 2023-01-30)

To quote [Wikipedia](https://en.wikipedia.org/wiki/Balmer_jump#:%7E:text=It%20is%20caused%20by%20electrons,wavelengths%20shorter%20than%20364.5%20nm.):

> Balmer Jump is caused by electrons being completely ionized directly from the second energy level of a hydrogen atom (bound-free absorption), which creates a continuum absorption at wavelengths shorter than 364.5 nm.

Based on the graphs below we used in class, the non-grey energy distribution of the energy flux is continuously lower than the grey case for wavelengths shorter than $364.5nm$, which according to Wikipedia is due to the bound-free absorptions of electrons in the $n=2$ energy state.

I understand why the energy flux gets lower for those wavelengths but I still cannot grasp why the discontinuity occurs. Is it because for wavelengths higher than $365.4nm$, the bound-free absorption of electrons of $n=3$ starts? And do bound-bound absorptions not occur at all for stars with these discontinuities?

I would be very thankful for any clarification on this subject.

[![enter image description here](https://i.sstatic.net/EWTtC.png)](https://i.sstatic.net/EWTtC.png)

**Tags:** star, spectra, stellar-atmospheres

> **Comments:**
> **username_1**: I wonder if there is any relationship to [Why does titanium oxide around Betelgeuse produce this particular sawtooth-shaped absorption spectrum?](https://astronomy.stackexchange.com/q/35451/7982)  
> **username_0**: Thank you for your comment.  Indeed the two spectra do have similar shapes, but reading the comments of the post mentioned it looks like the Betelgeuse's spectrum is due to the TiO behavior as a heavy metal, whereas the Balmer Jump is supposedly due to the bound-free transitions of n=2 Hydrogen electrons.  Thank you for the reference though, I will try to dig deeper that way as well.  

---

## 2 answers

### Answer by username_3 on 2023-02-01. Score: 4 | Accepted answer

The Balmer break comes from a combination of two main things: the ability of photons with high enough energies (wavelengths shorter than 364.5 nm) to ionize hydrogen atoms that are in the $n = 2$ energy level, and the *inability* of any photons with lower energies (wavelengths longer than 364.5 nm) to ionize the atoms. This leads to an abrupt change in absorption of outgoing photons, from high absorption at wavelengths shorter than 364.5 nm, to no absorption at longer wavelengths.

Another important factor: Any photon with a wavelength shorter than 364.5 nm can ionize an $n = 2$ hydrogen atom -- but the probability (due to the cross-section of the photon-atom interaction) goes *down* as the wavelength gets shorter. That's why you get the sawtooth pattern: very-short-wavelength photons *can* ionize the atoms, but are unlikely to. As the wavelengths get longer, the probability of ionizing a photon increases, up to the point of 364.5 nm. Once the wavelength is longer than that, ionization is simply impossible, so the absorption goes to zero.

In practice, the *total* absorption doesn't go to zero, because there are *other* sources of absorption, including hydrogen atoms in the higher states, like the $n = 3$ state (which has its own "Paschen jump"), and things like H$^{-}$ hydrogen ions. You also have to think about things like how many atoms are in different energy levels. In a very hot atmosphere, most of the hydrogen atoms will be in higher energy levels (or even ionized), so you won't get much of a Balmer break.

*do bound-bound absorptions not occur at all for stars with these discontinuities?*

They absolutely do occur -- they produce [*absorption lines*](https://en.wikipedia.org/wiki/Spectral_line), such as those of the [Balmer series](https://en.wikipedia.org/wiki/Balmer_series) (which are not shown in the simplified, cartoon spectra in your figure). If you have enough hydrogen atoms in the $n = 2$ state to produce a noticeable Balmer break, then you'll automatically have Balmer absorption lines (H$\alpha$, H$\beta$, etc.) due to longer-wavelength photons being absorbed by the same atoms.

These lines do get closer together as you approach 364.5 nm from the long-wavelength side, which has the effect of making the peak of the sawtooth a bit rounded, but they don't produce the jump -- they make it a little bit more gradual. (Your figures do not show the effects of the absorption lines.)

> **Comments:**
> **username_0**: Wow, this makes so much sense to me. Thank you very much for this detailed explanation.  

---

### Answer by username_2 on 2023-02-01. Score: 0

I don't quite understand your plot (does "grey" mean "blackbody"?), but I think you've stated what's going on pretty clearly (I'm reading *Fundamental Astronomy* by Karttumen et al to make sure).

The Balmer series starts with n=2 to n=3 at 656 nm, and as the final energy level increases, the lines get closer together ([I found this nifty calculator](https://www.vcalc.com/wiki/ekskekel/Balmer-Rydberg+Equation), which gives answers that look right).

$$n\_f=5\qquad \lambda=434~nm$$
$$n\_f=6\qquad \lambda=410~nm$$
$$n\_f=7\qquad \lambda=397~nm$$
$$...$$
$$n\_f=50,\qquad \lambda=365.091~nm$$
$$n\_f=51,\qquad \lambda =365.068~nm$$

And so forth, until the free state at 364.7 nm. Visually, this looks like some kind of continuum absorption, since the lines are not resolvable. But after that, you don't hit the next absorption line (Lyman) until 121.6 nm (that's n=1 to n=2). So calling that a jump (or discontinuity) is really more about the distribution of the lines - they go from being very close together to being regularly spaced apart again. Presumably we could also define "Lyman jump" at 91.2 nm, and Pachsen jump at 821 nm, etc.

> **Comments:**
> **username_3**: This isn't really correct. The "jump" is an *increase* in absorption at wavelengths *shorter* than 364.7 nm, due to photoionization of $n = 2$ atoms. The crowding together of individual absorption lines with higher and higher values of $n_{f}$ modifies the shape of the jump immediately longward of 364.7 nm, but that's all.  
> **username_2**: @username_3: Ah yeah thanks, that's closer to what the OP was asking, that explains his plots. The description in Karttuman et al is more about the crowding of the lines, but yeah, once you *can* ionize from the n=2 state, you'll get excess absorption because of it.  

---

