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Решение

A1. Stefan's law: \(L_{\odot}=\left(4 \pi R_{\odot}^{2}\right)\left(\sigma T_{\mathrm{s}}^{4}\right)\)\[T_{\mathrm{s}}=\left(\frac{L_{\odot}}{4 \pi R_{\odot}^{2} \sigma}\right)^{1 / 4}=5.76 \cdot 10^{3} \mathrm{~K}\]The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, \(u(\nu)\), is given by\[u(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} \nu^{3} \exp \left(-h \nu / k_{\mathrm{B}} T_{\mathrm{s}}\right),\]where \(A\) is the area of the surface normal to the direction of the incident radiation.

Now, consider a solar cell which consists of a thin disc of semiconducting material of area, \(A\), placed perpendicular to the direction of the Sun's rays.

Ответ:

\[T_{\mathrm{s}}=\left(\frac{L_{\odot}}{4 \pi R_{\odot}^{2} \sigma}\right)^{1 / 4}=5.76 \cdot 10^{3} \mathrm{~K}\]

A2. \[P_{\mathrm{in}}=\int_{0}^{\infty} u(\nu) d \nu=\int_{0}^{\infty} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} \nu^{3} \exp \left(-h \nu / k_{\mathrm{B}} T_{\mathrm{s}}\right) d \nu\] Let \(x=\dfrac{h \nu}{k_{\mathrm{B}} T_{\mathrm{s}}}\). Then, \(\nu=\dfrac{k_{\mathrm{B}} T_{\mathrm{s}}}{h} x \quad d \nu=\dfrac{k_{\mathrm{B}} T_{\mathrm{s}}}{h} d x\). \[P_{\text {in }}=\frac{2 \pi h A R_{\odot}^{2}}{c^{2} d_{\odot}^{2}} \frac{\left(k_{\mathrm{B}} T_{\mathrm{s}}\right)^{4}}{h^{4}} \int_{0}^{\infty} x^{3} e^{-x} d x=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \cdot 6=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}\]

Ответ:

\[P_{\text {in }}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}\]

A3. \[\begin{aligned} n_{\gamma}(\nu) & =\frac{u(\nu)}{h \nu} \\ & =A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} \exp \left(-h \nu / k_{\mathrm{B}} T_{\mathrm{s}}\right) \end{aligned}\] The semiconducting material of the solar cell has a "band gap" of energy, \(E_{\mathrm{g}}\). We assume the following model. Every photon of energy \(E \geq E_{\mathrm{g}}\) excites an electron across the band gap. This electron contributes an energy, \(E_{\mathrm{g}}\), as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).

Ответ:

\[n_{\gamma}(\nu) =A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} \exp \left(-h \nu / k_{\mathrm{B}} T_{\mathrm{s}}\right)\]

A4. The useful power output is the useful energy quantum per photon, \(E_{\mathrm{g}} \equiv h \nu_{\mathrm{g}}\), multiplied by the number of photons with energy, \(E \geq E_{\mathrm{g}}\).\[\begin{aligned} P_{\mathrm{out}} & =h \nu_{\mathrm{g}} \int_{\nu_{\mathrm{g}}}^{\infty} n_{\gamma}(\nu) d \nu \\ & =h \nu_{\mathrm{g}} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \int_{\nu_{\mathrm{g}}}^{\infty} \nu^{2} \exp \left(-h \nu / k_{\mathrm{B}} T_{\mathrm{s}}\right) d \nu \\ & =k_{\mathrm{B}} T_{\mathrm{s}} x_{\mathrm{g}} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}}\left(\frac{k_{\mathrm{B}} T_{\mathrm{s}}}{h}\right)^{3} \int_{x_{\mathrm{g}}}^{\infty} x^{2} e^{-x} d x \\ & =\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}} \end{aligned}\]

Ответ:

\[P_{\mathrm{out}}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}\]

A5. Efficiency \(\eta=\dfrac{P_{\text {out }}}{P_{\text {in }}}=\dfrac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}\)

\[\eta=\dfrac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}\]

A6. \[\eta=\frac{1}{6}\left(x_{\mathrm{g}}^{3}+2 x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}\right) e^{-x_{\mathrm{g}}}\]Put limiting values, \(\eta(0)=0 \quad \eta(\infty)=0\).\ Since the polynomial has all positive coefficients, it increases monotonically; the exponential function decreases monotonically. Therefore, \(\eta\) has only one maximum.\[\begin{aligned} & \frac{\mathrm{d} \eta}{\mathrm{~d} x_{\mathrm{g}}}=\frac{1}{6}\left(-x_{\mathrm{g}}^{3}+x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}} \\ & \left.\frac{\mathrm{~d} \eta}{\mathrm{~d} x_{\mathrm{g}}}\right|_{x_{\mathrm{g}}=0}=\left.\frac{1}{3} \quad \frac{\mathrm{~d} \eta}{\mathrm{~d} x_{\mathrm{g}}}\right|_{x_{\mathrm{g}} \rightarrow \infty}=0 \end{aligned}\]

Ответ:

A7. The maximum will be for \(\dfrac{\mathrm{d} \eta}{\mathrm{d} x_{\mathrm{g}}}=\dfrac{1}{6}\left(-x_{\mathrm{g}}^{3}+x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}=0\) \[\Rightarrow p\left(x_{\mathrm{g}}\right) \equiv x_{\mathrm{g}}^{3}-x_{\mathrm{g}}^{2}-2 x_{\mathrm{g}}-2=0\]A Numerical Solution by the Bisection Method:

\[2.25~{<}~x_{0}~{<}~2.5\]The approximate value of \(x_{\mathrm{g}}\) where \(\eta\) is maximum is \(x_{0}=2.27\). Alternative methods leading to the same result are acceptable.

Ответ:

\[\eta(2.27)=0.457\]

A8. \[\begin{gathered} x_{\mathrm{g}}=\dfrac{1.11 \cdot 1.60 \cdot 10^{-19}}{1.38 \cdot 10^{-23} \cdot 5763}=2.23 \\ \eta_{\mathrm{Si}}=\dfrac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}=0.457 \end{gathered}\]

In the late nineteenth century, Kelvin and Helmholtz (KH) proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, \(M_{\odot}\), and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational energy through this slow contraction.

Ответ:

\[\eta_{\mathrm{Si}}=0.457\]

A9. The total gravitational potential energy of the Sun: \[\Omega=-\int_{0}^{M_{\odot}} \frac{G m \mathrm{~d} m}{r}\] For constant density, \(\rho=\dfrac{3 M_{\odot}}{4 \pi R_{\odot}^{3}}, ~m=\frac{4}{3} \pi r^{3} \rho, ~\mathrm{~d} m=4 \pi r^{2} \rho \mathrm{~d} r\) \[\Omega=-\int_{0}^{R_{\odot}} G\left(\frac{4}{3} \pi r^{3} \rho\right)\left(4 \pi r^{2} \rho\right) \frac{\mathrm{d} r}{r}=-\frac{16 \pi^{2} G \rho^{2}}{3} \frac{R_{\odot}^{5}}{5}=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}\]

Ответ:

\[\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}\]

A10. \[\begin{gathered} \tau_{\mathrm{KH}}=\frac{-\Omega}{L_{\odot}} \\ \tau_{\mathrm{KH}}=\frac{3 G M_{\odot}^{2}}{5 R_{\odot} L_{\odot}}=1.88 \cdot 10^{7} ~\mathrm{years} \end{gathered}\] The \(\tau_{\mathrm{KH}}\) calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational.

Ответ:

\[\tau_{\mathrm{KH}}=1.88 \cdot 10^{7}~ \mathrm{years}\]

B1. \[4.0 \cdot 10^{-12} \mathrm{~J} \leftrightarrow 2 \nu\] \[\Phi_{\nu}=\frac{L_{\odot}}{4 \pi d_{\odot}^{2} \delta E} \cdot 2=\frac{3.85 \cdot 10^{26}}{4 \pi \cdot\left(1.50 \cdot 10^{11}\right)^{2} \cdot 4.0 \cdot 10^{-12}} \cdot2=6.8 \cdot 10^{14} \mathrm{~m}^{-2} \mathrm{~s}^{-1}\] Travelling from the core of the Sun to the Earth, some of the electron neutrinos, \(\nu_{\mathrm{e}}\), are converted to other types of neutrinos, \(\nu_{\mathrm{x}}\). The efficiency of the detector for detecting \(\nu_{\mathrm{x}}\) is \(1 / 6\) th of its efficiency for detecting \(\nu_{\mathrm{e}}\). If there is no neutrino conversion, we expect to detect an average of \(N_{1}\) neutrinos in a year. However, due to the conversion, an average of \(N_{2}\) neutrinos ( \(\nu_{\mathrm{e}}\) and \(\nu_{\mathrm{x}}\) combined) are actually detected per year.

\[\Phi_{\nu}=6.8 \cdot 10^{14} \mathrm{~m}^{-2} \mathrm{~s}^{-1}\]

B2. \[\begin{aligned} & N_{1}=\varepsilon N_{0} \\ & N_{e}=\varepsilon N_{0}(1-f) \\ & N_{x}=\varepsilon N_{0} f / 6 \\ & N_{2}=N_{e}+N_{x} \end{aligned}\] OR \[\begin{aligned} & (1-f) N_{1}+\frac{f}{6} N_{1}=N_{2} \\ & \Rightarrow f=\frac{6}{5}\left(1-\frac{N_{2}}{N_{1}}\right) \end{aligned}\] In order to detect neutrinos, large detectors filled with water are constructed. Although the interactions of neutrinos with matter are very rare, occasionally they knock out electrons from water molecules in the detector. These energetic electrons move through water at high speeds, emitting electromagnetic radiation in the process. As long as the speed of such an electron is greater than the speed of light in water (refractive index, \(n\) ), this radiation, called Cherenkov radiation, is emitted in the shape of a cone.

Ответ:

\[f=\frac{6}{5}\left(1-\frac{N_{2}}{N_{1}}\right)\]

B3. When the electron stops emitting Cherenkov radiation, its speed has reduced to \(v_{\text {stop }}=c / n\). Its total energy at this time is\[E_{\text {stop }}=\frac{m_{\mathrm{e}} c^{2}}{\sqrt{1-v_{\text {stop }}^{2} / c^{2}}}=\frac{n m_{\mathrm{e}} c^{2}}{\sqrt{n^{2}-1}}\]The energy of the electron when it was knocked out is\[E_{\mathrm{start}}=\alpha \Delta t+\frac{n m_{\mathrm{e}} c^{2}}{\sqrt{n^{2}-1}}\]Before interacting, the energy of the electron was equal to \(m_{\mathrm{e}} c^{2}\).

Thus, the energy imparted by the neutrino is\[E_{\text {imparted }}=E_{\text {start }}-m_{\mathrm{e}} c^{2}=\alpha \Delta t+\left(\frac{n}{\sqrt{n^{2}-1}}-1\right) m_{\mathrm{e}} c^{2}\]The fusion of H into He inside the Sun takes place in several steps. Nucleus of \({ }^{7} \mathrm{Be}\) (rest mass, \(m_{\mathrm{Be}}\) ) is produced in one of these intermediate steps. Subsequently, it can absorb an electron, producing a \({ }^{7} \mathrm{Li}\) nucleus (rest mass \(m_{\mathrm{Li}}<m_{\mathrm{Be}}\) ) and emitting a \(\nu_{\mathrm{e}}\). The corresponding nuclear reaction is:\[{ }^{7} \mathrm{Be}+\mathrm{e}^{-} \longrightarrow{ }^{7} \mathrm{Li}+\nu_{\mathrm{e}} .\]When a Be nucleus (\(m_{\mathrm{Be}}=11.65 \cdot 10^{-27} \mathrm{~kg}\)) is at rest and absorbs an electron also at rest, the emitted neutrino has energy \(E_{\nu}=1.44 \cdot10^{-13} \mathrm{~J}\). However, the Be nuclei are in random thermal motion due to the temperature \(T_{\mathrm{c}}\) at the core of the Sun, and act as moving neutrino sources. As a result, the energy of emitted neutrinos fluctuates with a root mean square value \(\Delta E_{\mathrm{rms}}\).

Ответ:

\[E_{\text {imparted }}=\alpha \Delta t+\left(\frac{n}{\sqrt{n^{2}-1}}-1\right) m_{\mathrm{e}} c^{2}\]

B4. Moving \({ }^{7}\mathrm{Be}\) nuclei give rise to Doppler effect for neutrinos. Since the fractional change in energy ( \(\Delta E_{\mathrm{rms}} / E_{\nu} \sim 10^{-4}\) ) is small, the Doppler shift may be considered in the nonrelativistic limit (a relativistic treatment gives almost same answer). Taking the line of sight along the \(z\)-direction, \[\begin{aligned} \frac{\Delta E_{\mathrm{rms}}}{E_{\nu}} & =\dfrac{v_{z, r m s}}{c} \\ & =3.85 \cdot 10^{-4} \\ & =\frac{1}{\sqrt{3}} \dfrac{V_{\mathrm{Be}}}{c} \end{aligned}\] \[\Rightarrow V_{\mathrm{Be}}=\sqrt{3} \cdot3.85 \cdot 10^{-4} \cdot 3.00 \cdot 10^{8} \mathrm{~m} \mathrm{~s}^{-1}=2.01 \cdot 10^{5} \mathrm{~m} \mathrm{~s}^{-1}.\] The average temperature is obtained by equating the average kinetic energy to the thermal energy. \[\begin{aligned} & \frac{1}{2} m_{\mathrm{Be}} V_{\mathrm{Be}}^{2}=\frac{3}{2} k_{\mathrm{B}} T_{\mathrm{c}}  \end{aligned}\]

Ответ:

\[T_{\mathrm{c}}=1.13 \cdot 10^{7} \mathrm{~K}\]

Перевод: pho.rs.