IOM 2020, Теория — Задача 3. PLANETARY NEBULA

• Part I: Star radiation and gas temperature in the nebula., 1) $\sigma = \frac{2\pi^5k^4}{15c^2h^3}$

• Part I: Star radiation and gas temperature in the nebula., 2) $\frac{T(r)}{T_0} = \sqrt{\frac{r_0}{r}}$

• Part II: Hydrogen atom., 3) $E_n = -\frac{m_ee^4}{8\varepsilon_0^2h^2n^2}$

• Part II: Hydrogen atom., 4) 13.6 eV

• Part II: Hydrogen atom., 5) n = 1 (the ground state); practically no excited atoms: 0 %

  Official solution: T(r) ≈ 95 K, the probability of excitation is of the order of e^(−ΔE₂₁/kT) ≈ 10⁻⁵⁴¹.

• Part II: Hydrogen atom., 6) 20 %

  Official solution: δ ≈ 0,194 ± 0,015; a reasonable answer is δ ~ 20 %.

• Part III. Radiation of nebula., 7) $1.7 \times 10^{13}$ cm

• Part III. Radiation of nebula., 8) 100 %

• Part III. Radiation of nebula., 9) ΔN₂₁ : ΔN_B : ΔN ≈ 1 : 1 : 1

• Part III. Radiation of nebula., 10) Official sketch: the Planck spectrum of the star cut off at ν_i = E_i/h, a strong line of the main Lyman series at ν₂₁, weaker Balmer lines and lines of other series at lower frequencies.

• Part III. Radiation of nebula., 11) $\frac{1{,}5z(2z+1)^2e^{-z/2}}{(z+1)^3e^{z/2} - (1{,}5z+1)^3} = 0{,}1,\quad z = \frac{2}{3}\frac{h\nu_{21}}{kT_x} \approx \frac{79000\ \mathrm{K}}{T_x}$ ≈ $19250$ K

  Official solution: z ≈ 4,1, T_x ≈ 19250 K.

• Part III. Radiation of nebula., 12) l₁ ≈ 1,6·10¹⁵ cm; l₂ ≈ 1,8·10¹⁵ cm; l₃ ≈ 1,9·10¹⁵ cm

Proposed solution and answers

1. The total density of radiation being in equilibrium with a surface of hot body is calculated by integrating over the whole spectrum: $W = \displaystyle\int_0^{\infty}\frac{8\pi h}{c^3}\frac{\nu^3}{e^{h\nu/kT} - 1}d\nu = \frac{8\pi h}{c^3}\left(\frac{kT}{h}\right)^4\int_0^{\infty}dx\frac{x^3}{e^x - 1} = \frac{8\pi^5}{15}\frac{k^4T^4}{h^3c^3}$. Here the integral given in problem statement is used. To determine the energy flux being emitted by the surface note that «equilibrium» radiation propagates at the speed of light from every point uniformly in all directions. Therefore, any element of solid angle $d\Omega$ emits the energy flux of $dJ = \dfrac{cW}{4\pi}d\Omega$. Let $\vartheta$ be an angle counted from a surface normal. The outgoing radiation is emitted into the angular interval $0 \le \vartheta \le \dfrac{\pi}{2}$ and an infinitesimal solid angle between conic surfaces with apex angles of $\vartheta$ and $\vartheta + d\vartheta$ equals $d\Omega = 2\pi\sin\vartheta\,d\vartheta$. Then the total energy flux perpendicular to the surface is $J = 2\pi\dfrac{cW}{4\pi}\displaystyle\int_0^{\pi/2}\sin\vartheta\cos\vartheta\,d\vartheta = \frac{cW}{4} = \frac{2\pi^5k^4}{15c^2h^3}T^4$. Comparing this equation with the Stefan-Boltzmann law one obtains: $\sigma = \dfrac{2\pi^5k^4}{15c^2h^3}$.

2. In equilibrium nebula the energy flux emitted by the star is constant and equals to the flux from the star surface. If a gas layer at a distance $r \gg r_0$ from the star were in thermal equilibrium with the radiation it would emit to «outside» the same flux as received from the «inside». According to the Stefan-Boltzmann law the flux temperature would be determined by the equation: $\sigma\cdot T^4(r)\cdot 4\pi r^2 = \sigma\cdot T_0^4\cdot 4\pi r_0^2$. Hence, $\dfrac{T(r)}{T_0} = \sqrt{\dfrac{r_0}{r}}$.

3. For a circular orbit the Bohr model requires $\dfrac{2\pi r}{\lambda} = n\ (n \in Z)$, so $r\cdot p = m_evr = \dfrac{h}{2\pi}\cdot n$. Writing down the classical equation of motion of electron in a circular orbit of a radius $r$ at a velocity $v$, one obtains for hydrogen atom: $m_e\dfrac{v^2}{r} = \dfrac{e^2}{4\pi\varepsilon_0r^2}$, then $r = \dfrac{e^2}{4\pi\varepsilon_0m_ev^2}$ and $m_evr = \dfrac{e^2}{4\pi\varepsilon_0v} = n\dfrac{h}{2\pi}$. Therefore, the velocity and the radius of electron orbit are $v = \dfrac{e^2}{2\varepsilon_0hn}$, $r = \dfrac{\varepsilon_0h^2n^2}{\pi m_ee^2}$ and the energy of electron in a stationary orbit is $E_n = \dfrac{m_ev^2}{2} - \dfrac{e^2}{4\pi\varepsilon_0r} = -\dfrac{m_ee^4}{8\varepsilon_0^2h^2n^2}$.

4. To ionize atom, its electron must be moved far away from the nucleus, where $E = 0$, i.e. to move the electron from the state with $n = 1$ to a state with $n \to \infty$. Therefore, the ionization threshold of hydrogen atom is $E_i = -E_1 = \dfrac{m_ee^4}{8\varepsilon_0^2h^2} \approx 13{,}6$ eV.

5. According to 2), the «equilibrium» gas temperature at $r = 10^5r_0$ turns out to be $T(r) = T_0\sqrt{\dfrac{r_0}{r}} \approx 95$ K. The energy of chaotic thermal motion of hydrogen atoms at this temperature is $\dfrac{3}{2}kT \approx 0{,}012$ eV. The energy required to move electron to the lowest excited state (with $n$ = 2) is $\Delta E_{21} = E_2 - E_1 = \dfrac{3m_ee^4}{32\varepsilon_0^2h^2} = \dfrac{3}{4}E_i \approx 10{,}2$ eV, i.e. a thousand times more! It should be obvious that a probability of transition to an excited state is very small, practically vanishing – evidently less than one percent! The contestants familiar with the Boltzmann distribution can evaluate this probability more precisely: it is of the order of $e^{-\Delta E_{21}/kT} \approx 10^{-541}$. That is, there would be practically no excited atoms at a given distance from the star.

6. The photons able to ionize hydrogen atoms must have an energy of $h\nu \ge h\nu_i = E_i \approx 13{,}6$ eV. For $T_0 = 30000$ K this yields $\varepsilon \equiv \dfrac{h\nu_i}{kT_0} = \dfrac{E_i}{kT_0} \approx 5{,}28$. The energy distribution of photons is given by the Planck distribution; therefore, the desired photon fraction is $\delta = \left(\displaystyle\int_{\nu_i}^{\infty}w(\nu)d\nu\right)\Big/\left(\displaystyle\int_0^{\infty}w(\nu)d\nu\right)$. Using the information from the problem statement yields: $\delta = \dfrac{15}{\pi^4}\displaystyle\int_{\varepsilon}^{\infty}\frac{x^3}{e^x - 1}dx \approx \frac{15}{\pi^4}(\varepsilon + 1)^3e^{-\varepsilon} \approx 0{,}194 \pm 0{,}015$. A reasonable answer is $\delta \sim 20$ %.

7. Density of hydrogen atoms at a mean mass density $\bar{\rho}$ is $\bar{n} = \dfrac{\bar{\rho}}{m_H}$, where $m_H$ is the mass of a single atom. This mass can be expressed via the Avogadro constant: obviously, $m_H = \dfrac{1\ \mathrm{g/mole}}{N_A} \approx 1{,}7\cdot 10^{-24}$ g. Thus, $\bar{n} \approx 6\cdot 10^3$ см$^{-3}$. To get an estimate one can assume that every atom creates a photon «trap» with a cross section $\sigma_i$. Then the mean free path of a photon is determined by the requirement that these «traps» overlap the entire flux cross section: $\bar{n}\cdot S\cdot l_{\gamma} = \dfrac{S}{\sigma_i}$, whence $l_{\gamma} = \dfrac{1}{\sigma_i\bar{n}} \approx 1{,}7\cdot 10^{13}$ cm.

8. One can see that the nebula radius is about 6 thousand times greater than the «mean» free path of a photon. This means that practically all photons (all 100%, when rounded to an integer) with an energy exceeding the ionization threshold will be absorbed by the atoms and converted into photons of lower energies. (Note that according to 6), $\varepsilon \equiv \dfrac{E_i}{kT_0} \approx 5{,}28$, i.e. the ionization energy is higher than the maximum $E_m \approx 2{,}84\cdot kT_0$ of the photon energy distribution. So, as the energy of the photons grows, their contribution to the total flux rapidly decreases and the energy dependence of the ionization cross section could not significantly affect the result of the estimate.)

9. According to 8), almost all photons with energies exceeding the ionization threshold are being absorbed by the nebula. Each of the absorbed photons triggers a chain of transitions. If the first transition occurred to the state $n = 1$, then a photon of Lyman series was emitted, and this photon almost certainly would be absorbed by another hydrogen atom (the majority of hydrogen atoms of the nebula are in the ground state, i.e. the energy of a Lyman photon corresponds to an allowed transition for them and the mean free path is much less than the nebula radius). The atom that had absorbed this photon would return in the same excited state. This would continue until a transition to state with $n \ne 1$ occurs. If the first transition after the capture occurred in a state with $n = 2$, then a Balmer series photon would be emitted which would be inevitably followed by emission of a photon of the main Lyman series with a frequency $\nu_{21}$. Balmer photons are practically not absorbed (since a fraction of hydrogen atoms in a state with $n = 2$ is extremely low in the nebula) and leave the nebula after many elastic scattering events. Photons of the main Lyman series are absorbed and reemitted many times before leaving the nebula as well. If the first transition occured to a level with $n > 2$ the situation is «repeated»: there is again a «highly excited» state of atom which would follow a transition in the state $n = 1$, or in a state with $n = 2$, or in a state with $n > 2$. Therefore, every absorbed photon of high energy necessarily will «parent» a single Balmer photon and a single photon of the main Lyman series (plus several photons of other series of lesser energies). The contribution of Balmer photons and photons of the main Lyman series «originally» emitted by the star is very small due to a small width of the line and a finite density of thermal radiation. Thus, $\Delta N_{21} : \Delta N_B : \Delta N \approx 1:1:1$.

10. The spectrum of nebula radiation observed by an external observer is the spectrum of star radiation practically coinciding with the Planck spectrum corresponding to its temperature and «reprocessed» by the nebula. According to 8 and 9, the thermal spectrum will be practically «cut» from above at the frequency $\nu_i = \dfrac{E_i}{h}$ due to the almost complete absorption of the photons with energies above the ionization threshold. The energy emitted by this part of the spectrum is mostly transformed into the energy of photons of the main Lyman series (the energy of this photon is up to 75% of the energy of the «processed» photon), less energy goes to photons of Balmer series, and even less to low energy photons of other series. Therefore, the typical spectrum will have the approximate shape shown in the Figure.

11. In the given frequency interval $\Delta\nu$ there are no spectral lines of hydrogen (except for the Lyman line which contribution to the intensity is not counted), so the intensity in this frequency interval is determined by the «unprocessed» star radiation.

Here $z \equiv \dfrac{2}{3}\dfrac{h\nu_{21}}{kT_x} \approx \dfrac{79000\ \mathrm{K}}{T_x}$, and the numerical factor was written down by taking into account that the power radiated from a unit area is related to the spectral density as $J = \dfrac{cW}{4}$ (see 1). Note that an accurate evaluation of the factor is not necessary since it cancels out anyway. The radiation intensity at the frequency $\nu_{21}$ is determined by the condition that the number of photons with this frequency coming out of the nebula equals the number of photons with energies above the ionization threshold. Therefore,

This calculation takes into account that $\nu_i = \dfrac{4}{3}\nu_{21} = 2\cdot\dfrac{2}{3}\nu_{21} \Rightarrow \dfrac{h\nu_i}{kT_0} = 2z$. According to the observer’s data, $\dfrac{I(\nu_{21})}{I(\Delta\nu)} = \dfrac{1{,}5z(2z + 1)^2e^{-z/2}}{(z + 1)^3e^{z/2} - (1{,}5z + 1)^3} = 0{,}1$. This is the equation for $z$, whence $T_x \approx \dfrac{79000\ \mathrm{K}}{z}$.

This equation has to be numerically solved by evaluating the values of $f(z) \equiv \dfrac{1{,}5z(2z + 1)^2e^{-z/2}}{(z + 1)^3e^{z/2} - (1{,}5z + 1)^3}$ for various $z$ and comparing with 0.1 (it would me more effective to use the method of dichotomy or, if available, to use Excel or a programmable calculator and simply look at the values of $f(z)$ in the «expected» range):

One can see, the «appropriate» value is $z \approx 4{,}1$. Therefore, $T_x \approx 19250$ K.

12. The equilibrium degree of ionization is achieved when the number of ionization events in a «thin layer» of gas per unit of time (equal to the number of absorbed photons with energies above the ionization threshold) equals the number of «captures» of electrons by protons (recombinations). The flux of such photons decreases as the distance from the star increases not only because the surface area grows but also because of absorption: $I(r)\cdot 4\pi r^2 - I(r + dr)\cdot 4\pi(r + dr)^2 = I(r)\sigma_in_1\cdot 4\pi r^2dr$. Here $n_1 = n_1(r)$ is the density of hydrogen atoms absorbing such photons (not ionized). Thus, $\dfrac{dI}{dr} = -\left(\dfrac{2}{r} + \sigma_in_1\right)I \approx -\sigma_in_1I$, in accordance with the assumptions of the model used. Integration of this equation yields: $I(r) \approx I_0\cdot\exp\left(-\sigma_i\displaystyle\int_{r_0}^rn_1dr\right)$. The number of photons absorbed per unit time is proportional to $\sigma_in_1I(r)$. The number of captures is proportional to the product of electron density $n_e$ and the density of hydrogen ions $n_i$ which are equal because the ionized layer remains neutral. Thus, $n_1\cdot\exp\left(-\sigma_i\displaystyle\int_{r_0}^rn_1dr\right) = const\cdot n_e\cdot n_i = const\cdot n_i^2$. If the ionization degree is $\dfrac{n_i}{n} \equiv y$, then $n_1 = (1 - y)n$. According to the problem statement the ionization degree varies significantly faster than the concentration of atoms, so $\dfrac{y^2}{1 - y} \approx const\cdot\exp\left(-\dfrac{1}{l_{\gamma}}\displaystyle\int_{r_0}^r(1 - y)dr\right)$, where $l_{\gamma} = \dfrac{1}{\sigma_i\bar{n}}$ is the mean free path of photons evaluated in 7. Taking the logarithm of this equation and differentiating the result, one obtains: $\left(\dfrac{2}{y} + \dfrac{1}{1 - y}\right)dy = -(1 - y)\dfrac{dr}{l_{\gamma}}$, i.e. $dr = -l_{\gamma}\left(\dfrac{2}{y} + \dfrac{1}{1 - y}\right)\dfrac{dy}{1 - y}$. Integrating the latter from the star surface to a given radius, one obtains: $r - r_0 = l_{\gamma}\cdot\left[2\ln\left(\dfrac{y_0(1 - y)}{(1 - y_0)y}\right) + \dfrac{1}{1 - y_0} - \dfrac{1}{1 - y}\right]$. Here $y_0 \approx 0{,}99$ is the given degree of ionization near the star surface. Using this formula, we find: $l_1 \approx l_{\gamma}[90 + 2\ln(11)] \approx 1{,}6\cdot 10^{15}$ cm, $l_2 \approx l_{\gamma}[98 + 2\ln(99)] \approx 1{,}8\cdot 10^{15}$ cm, and $l_3 \approx l_{\gamma}\left[\dfrac{890}{9} + 2\ln(891)\right] \approx 1{,}9\cdot 10^{15}$ cm. One can see that the ionization degree becomes less than 1% at a distance of $l = 2\cdot 10^{15}$ cm! That is, the degree of ionization remains very high at a distance of about ~1 % of the nebula radius ($10^{15}$ cm), and drops almost to zero at a distance of ~2 %. Such a behavior confirms the assumption used in our model that the degree of ionization varies much faster than the area of the spherical surface grows and the density of the cloud decreases. Actually, it is pretty obvious that the concentration of atoms near the star is still higher than in the nebula periphery, so in the region with $r < 2\cdot 10^{15}$ m the mean free path of photons is less than the «average» value used in our calculations. Therefore, in reality the region of high ionization is even stronger «pressed» against the star.