IChO 2006 · Theory problem 2

Detection of hydrogen

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Hydrogen is prevalent in the universe. Life in the universe is ultimately based on hydrogen.

2.1 There are about \(1\times10^{23}\) stars in the universe. Assume that they are like our sun (radius, 700,000 km; density, \(1.4\ \mathrm{g\ cm^{-3}}\); 3/4 hydrogen and 1/4 helium by mass). Estimate the number of stellar protons in the universe to one significant figure.

In the 1920s Cecilia Payne discovered by spectral analysis of starlight that hydrogen is the most abundant element in most stars.

2.2 The electronic transition of a hydrogen atom is governed by \(\Delta E(n_i\to n_f) =\)
\(= -C(1/n_f^2-1/n_i^2)\), where \(n\) is principle quantum number, and \(C\) is a constant. For detection of the \(\Delta E(3\to2)\) transition (656.3 nm in the Balmer series), the electron in the ground state of the hydrogen atom needs to be excited first to the \(n=2\) state. Calculate the wavelength (in nm) of the absorption line in the starlight corresponding to the \(\Delta E(1\to2)\) transition.

2.3 According to Wien's law, the wavelength (\(\lambda\)) corresponding to the maximum light intensity emitted from a blackbody at temperature \(T\) is given by \(\lambda T=2.9\times10^{-3}\ \mathrm{m\ K}\). Calculate the surface temperature of a star whose blackbody radiation has a peak intensity corresponding to the \(n=1\to n=2\) excitation of hydrogen.

The ground state of hydrogen is split into two hyperfine levels due to the interaction between the magnetic moment of the proton and that of the electron. In 1951, Purcell discovered a spectral line at 1420 MHz due to the hyperfine transition of hydrogen in interstellar space.

2.4 Hydrogen in interstellar space cannot be excited electronically by starlight. However, the cosmic background radiation, equivalent to 2.7 K, can cause the hyperfine transition. Calculate the temperature of a blackbody whose peak intensity corresponds to the 1420 MHz transition.

2.5 Wien generated hydrogen ions by discharge of hydrogen gas at a very low pressure and determined the \(e/m\) value, which turned out to be the highest among different gases tested. In 1919, Rutherford bombarded nitrogen with alpha-particles and observed emission of a positively charged particle which turned out to be the hydrogen ion observed by Wien. Rutherford named this particle the “proton”. Fill in the blank.

\(\underline{\hspace{4cm}}\)

Official English text of the problem.

Answers

  • 2.1: \(\boxed{1\times10^{80}\text{ stellar protons}}\), including protons within helium nuclei.
  • 2.2: \(\boxed{121.5\ \mathrm{nm}}\), the Lyman-\(\alpha\) absorption wavelength.
  • 2.3: \(\boxed{2.4\times10^{4}\ \mathrm{K}}\).
  • 2.4: \(\boxed{1.4\times10^{-2}\ \mathrm{K}}\), using the supplied wavelength form of Wien’s law.
  • 2.5: \(\boxed{\ce{^{14}_{7}N + ^{4}_{2}He -> ^{17}_{8}O + ^{1}_{1}H}}\). The emitted particle is a proton, the hydrogen-1 cation, [1H+].

Solution

2.1

For this estimate, take hydrogen to be hydrogen-1 and helium to be helium-4, with molar masses approximately \(1\ \mathrm{g\,mol^{-1}}\) and \(4\ \mathrm{g\,mol^{-1}}\), respectively. Use \(N_{\mathrm A}=6.022\times10^{23}\ \mathrm{mol^{-1}}\).

The radius of each star is

\[ R=700{,}000\ \mathrm{km}=7.00\times10^{10}\ \mathrm{cm}. \]

Its mass is therefore

\[ \begin{aligned} M_\star &=\rho\frac{4\pi R^3}{3}\\ &=(1.4\ \mathrm{g\,cm^{-3}}) \frac{4\pi}{3}(7.00\times10^{10}\ \mathrm{cm})^3\\ &=2.011\times10^{33}\ \mathrm{g}. \end{aligned} \]

Each hydrogen nucleus contains one proton, and each helium nucleus contains two. Thus the number of protons per star is

\[ \begin{aligned} N_{p,\star} &= \frac{(3/4)M_\star}{1\ \mathrm{g\,mol^{-1}}}N_{\mathrm A} + 2\frac{(1/4)M_\star}{4\ \mathrm{g\,mol^{-1}}}N_{\mathrm A}\\ &= (2.011\times10^{33}\ \mathrm{g}) \left(0.75+0.125\right) \frac{6.022\times10^{23}}{\mathrm{g}}\\ &=1.060\times10^{57}. \end{aligned} \]

Multiplying by the number of stars gives

\[ N_{p,\mathrm{universe}} =(1\times10^{23})(1.060\times10^{57}) =1.060\times10^{80}. \]

To one significant figure,

\[ \boxed{N_{p,\mathrm{universe}}=1\times10^{80}}. \]

2.2

The energy change of the atom in the \(3\to2\) emission transition is

\[ \Delta E_{3\to2} =-C\left(\frac14-\frac19\right) =-\frac{5C}{36}. \]

The emitted photon therefore has positive energy

\[ \frac{hc}{\lambda_{32}}=\frac{5C}{36}. \]

For absorption from \(n=1\) to \(n=2\),

\[ \Delta E_{1\to2} =-C\left(\frac14-1\right) =\frac{3C}{4} =\frac{hc}{\lambda_{12}}. \]

Taking the ratio eliminates \(h\), \(c\), and \(C\):

\[ \frac{\lambda_{32}}{\lambda_{12}} =\frac{3/4}{5/36} =\frac{27}{5}. \]

Hence

\[ \lambda_{12} =\frac{5}{27}(656.3\ \mathrm{nm}) =121.537\ldots\ \mathrm{nm}, \]

so

\[ \boxed{\lambda_{12}=121.5\ \mathrm{nm}}. \]

This is the Lyman-\(\alpha\) line in the ultraviolet.

2.3

Using the wavelength found in 2.2 and the supplied Wien constant,

\[ T=\frac{2.9\times10^{-3}\ \mathrm{m\,K}} {121.537\times10^{-9}\ \mathrm{m}} =2.3861\times10^{4}\ \mathrm{K}. \]

The Wien constant has two significant figures, giving

\[ \boxed{T=2.4\times10^{4}\ \mathrm{K}}. \]

2.4

First convert the transition frequency to a wavelength, using \(c=2.998\times10^8\ \mathrm{m\,s^{-1}}\):

\[ \nu=1420\ \mathrm{MHz}=1.420\times10^9\ \mathrm{s^{-1}}, \]

\[ \lambda=\frac{c}{\nu} =\frac{2.998\times10^8\ \mathrm{m\,s^{-1}}} {1.420\times10^9\ \mathrm{s^{-1}}} =0.2111\ \mathrm{m}. \]

This is the approximately \(21\ \mathrm{cm}\) hydrogen line. Applying the wavelength form of Wien’s law supplied in 2.3,

\[ T=\frac{2.9\times10^{-3}\ \mathrm{m\,K}} {0.2111\ \mathrm{m}} =1.374\times10^{-2}\ \mathrm{K}. \]

Therefore,

\[ \boxed{T=1.4\times10^{-2}\ \mathrm{K}}. \]

The \(2.7\ \mathrm{K}\) background can still supply photons at this frequency because a blackbody emits over a continuous range of wavelengths.

2.5

The incident alpha-particle is a helium-4 nucleus, [4He+2]. The emitted proton is a hydrogen-1 cation, [1H+].

Taking the nitrogen target to be nitrogen-14, write the reaction with an unknown residual nucleus:

\[ \ce{^{14}_{7}N + ^{4}_{2}He -> ^{A}_{Z}X + ^{1}_{1}H}. \]

Conservation of nucleon number gives

\[ 14+4=A+1 \quad\Rightarrow\quad A=17. \]

Conservation of nuclear charge gives

\[ 7+2=Z+1 \quad\Rightarrow\quad Z=8. \]

Atomic number 8 identifies oxygen, so the residual nucleus is oxygen-17. The completed reaction is

\[ \boxed{\ce{^{14}_{7}N + ^{4}_{2}He -> ^{17}_{8}O + ^{1}_{1}H}}. \]

Both sides contain 18 nucleons and total nuclear charge \(+9e\). The equation describes a nuclear transformation; electron shells and chemical bonding are not specified.

The solution was checked against the official solution.