3.3 Calculate the energy of the ion using the value of the lowest energy. · OlympiadBench #1223

Physics · modern-physics

Problem

3.3 Calculate the energy of the ion A(Z−1)+\mathrm{A}^{(\mathrm{Z}-1)+} using the value of the lowest energy. The calculation should be approximated based on the following principles: 3.3.A The potential energy of the ion should be expressed in terms of the average value of 1r\frac{1}{r}. (ie. 1r0;r0\frac{1}{\mathrm{r}_{\mathrm{0}}} ; \mathrm{r}_{0} is given in the problem). 3.3.B In calculating the kinetic energy of the ion, use the average value of the square of the momentum given in 3.1 after being simplified by (p0)2⋅(r0)2≈(ℏ)2\left(\mathrm{p}_{\mathrm{0}}\right)^{2} \cdot\left(\mathrm{r}_{\mathrm{0}}\right)^{2} \approx(\hbar)^{2} Given constants: Recombination of Positive and Negative Ions in Ionized Gas Introduction A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol A(Z−1)+\mathrm{A}^{(\mathrm{Z}-1)+} Constants:
electric field constantε0=8.85⋅10−12A⋅sV⋅m\varepsilon_{\mathrm{0}}=8.85 \cdot 10^{-12} \frac{\mathrm{A} \cdot \mathrm{s}}{\mathrm{V} \cdot \mathrm{m}}
elementary chargee=±1.602⋅10−19 A⋅s\mathrm{e}= \pm 1.602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{s}
q2=e24⋅π⋅ε0=2.037⋅10−28 J⋅m\mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{\mathrm{0}}}=2.037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{m}
Planck's constantℏ=1.054⋅10−34 J⋅s\hbar=1.054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}
(rest) mass of an electronme=9.108⋅10−31 kg\mathrm{m}_{\mathrm{e}}=9.108 \cdot 10^{-31} \mathrm{~kg}
Bohr's atomic radiusrB=ℏm⋅q2=5.92⋅10−11 m\mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5.92 \cdot 10^{-11} \mathrm{~m}
Rydberg's energyER=q22⋅rB=2.180⋅10−18 J\mathrm{E}_{\mathrm{R}}=\frac{\mathrm{q}^{2}}{2 \cdot \mathrm{r}_{\mathrm{B}}}=2.180 \cdot 10^{-18} \mathrm{~J}
(rest) mass of a protonmP⋅c2=1.503⋅10−10 J\mathrm{m}_{\mathrm{P}} \cdot \mathrm{c}^{2}=1.503 \cdot 10^{-10} \mathrm{~J}
Context question: 3.1 Assume that the ion which has just one electron left the shell. A(Z−1)+\mathrm{A}^{(\mathrm{Z}-1)+} is in the ground state. In the lowest energy state, the square of the average distance of the electron from the nucleus or r2\mathrm{r}^{2} with components along x−y\mathrm{x}-\mathrm{y} - and z\mathrm{z}-axis being (Δx)2,(Δy)2(\Delta \mathrm{x})^{2},(\Delta \mathrm{y})^{2} and (Δz)2(\Delta \mathrm{z})^{2} respectively and r02=(Δx)2+(Δy)2+(Δz)2\mathrm{r}_{\mathrm{0}}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2} \quad and also the square of the average momentum by po2=(Δpx)2+(Δpy)2+(Δpz)2\mathrm{p}_{\mathrm{o}}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}, whereas Δpx≥ℏ2⋅Δx,Δpy≥ℏ2⋅Δy\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}}, \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}} and Δpz≥ℏ2⋅Δz\Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}. Write inequality involving (p0)2⋅(r0)2\left(p_{\mathrm{0}}\right)^{2} \cdot\left(r_{\mathrm{0}}\right)^{2} in a complete form. Context answer: p02⋅r02≥94⋅ℏ2\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2} Context question: 3.2 The ion represented by A(Z−1)+\mathrm{A}^{(\mathrm{Z}-1)+} may capture an additional electron and consequently emits a photon. Write down an equation which is to be used for calculation the frequency of an emitted photon. Context answer: 3.2 ∣v⃗e∣…..\left|\vec{v}_{\mathrm{e}}\right| \ldots . .. speed of the external electron before the capture ∣V⃗i∣…..\left|\vec{V}_{i}\right| \ldots . .. speed of A(Z−1)+A^{(Z-1)+} before capturing ∣V⃗f∣…..\left|\vec{V}_{\mathrm{f}}\right| \ldots . .. speed of A(Z−1)+\mathrm{A}^{(\mathrm{Z}-1)+} after capturing En=h.v…..\mathrm{E}_{\mathrm{n}}=\mathrm{h} . v \quad \ldots . .. energy of the emitted photon conservation of energy: 12⋅me⋅Ve2+12⋅(M+me)⋅Vi2+E[A(Z−1)+]=12⋅(M+2⋅me)⋅Vf2+E[A(Z−2)+]\frac{1}{2} \cdot m_{e} \cdot V_{e}^{2}+\frac{1}{2} \cdot\left(M+m_{e}\right) \cdot V_{i}^{2}+E\left[A^{(Z-1)+}\right]=\frac{1}{2} \cdot\left(M+2 \cdot m_{e}\right) \cdot V_{f}^{2}+E\left[A^{(Z-2)+}\right] where E[A(Z−1)+)\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-1)+}\right) and E[A(Z−2)+]\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-2)+}\right] denotes the energy of the electron in the outermost shell of ions A(Z−1)+\mathrm{A}^{(\mathrm{Z}-1)+} and A(Z−2)+\mathrm{A}^{(\mathrm{Z}-2)+} respectively. conservation of momentum: me⋅v⃗e+(M+m)⋅v⃗i=(M+2⋅me)⋅v⃗f+h⋅vc⋅1→m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1} where 1→\overrightarrow{1} is the unit vector pointing in the direction of the motion of the emitted photon.

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