Physics

Physics · modern-physics

Problem

A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, Ω\Omega, of the Sun at present, in terms of G,M⊙G, M_{\odot} and R⊙R_{\odot}. Given constants: Particles from the Sun 1{ }^{1} Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be M⊙=2.00×1030 kgM_{\odot}=2.00 \times 10^{30} \mathrm{~kg}, its radius, R⊙=7.00×R_{\odot}=7.00 \times 108 m10^{8} \mathrm{~m}, its luminosity (radiation energy emitted per unit time), L⊙=3.85×1026 WL_{\odot}=3.85 \times 10^{26} \mathrm{~W}, and the Earth-Sun distance, d⊙=1.50×1011 md_{\odot}=1.50 \times 10^{11} \mathrm{~m}. Note: (i) ∫xeaxdx=(xa−1a2)eax+\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+ constant (ii) ∫x2eaxdx=(x2a−2xa2+2a3)eax+\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+ constant (iii) ∫x3eaxdx=(x3a−3x2a2+6xa3−6a4)eax+\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, TsT_{\mathrm{s}}, of the solar surface. Context answer: 5.76×1035.76 \times 10^{3} Extra Supplementary Reading Materials: 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(v)u(v), is given by u(v)=AR⊙2d⊙22πhc2v3exp⁡(−hv/kBTs)u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) where vv is the frequency and AA 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, AA, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, PinP_{\mathrm{in}}, incident on the surface of the solar cell, in terms of A,R⊙,d⊙,TSA, R_{\odot}, d_{\odot}, T_{\mathrm{S}} and the fundamental constants c,h,kBc, h, k_{\mathrm{B}} Context answer: Pin=12πkB4c2h3Ts4AR⊙2d⊙2P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} Context question: A3 Express the number of photons, nγ(v)n_{\gamma}(v), per unit time per unit frequency interval incident on the surface of the solar cell in terms of A,R⊙,d⊙,Ts,vA, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v and the fundamental constants c,h,kBc, h, k_{\mathrm{B}}. Context answer: nγ(ν)=AR⊙2d⊙22πc2ν2e−hνkBTsn_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a "band gap" of energy, EgE_{\mathrm{g}}. We assume the following model. Every photon of energy E≥EgE \geq E_{\mathrm{g}} excites an electron across the band gap. This electron contributes an energy, EgE_{\mathrm{g}}, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define xg=hvg/kBTsx_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}} where Eg=hvgE_{\mathrm{g}}=h v_{\mathrm{g}}. Express the useful output power of the cell, Pout P_{\text {out }}, in terms of xg,Ax_{\mathrm{g}}, A, R⊙,d⊙,TsR_{\odot}, d_{\odot}, T_{\mathrm{s}} and the fundamental constants c,h,kBc, h, k_{\mathrm{B}}. Context answer: Pout =2πkB4c2h3Ts4AR⊙2d⊙2xg(xg2+2xg+2)e−xgP_{\text {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}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}} Context question: A5 Express the efficiency, η\eta, of this solar cell in terms of xgx_{\mathrm{g}}. Context answer: η=xg6(xg2+2xg+2)e−xg\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}} Context question: A6 Make a qualitative sketch of η\eta versus xgx_{\mathrm{g}}. The values at xg=0x_{\mathrm{g}}=0 and xg→∞x_{\mathrm{g}} \rightarrow \infty should be clearly shown. What is the slope of η(xg)\eta\left(x_{\mathrm{g}}\right) at xg=0x_{\mathrm{g}}=0 and xg→∞x_{\mathrm{g}} \rightarrow \infty ? Context answer: 13\frac{1}{3} , 0 Context question: A7 Let x0x_{0} be the value of xgx_{\mathrm{g}} for which η\eta is maximum. Obtain the cubic equation that gives x0x_{0}. Estimate the value of x0x_{0} within an accuracy of \pm 0.25. Hence calculate η(x0)\eta\left(x_{0}\right). Context answer: x0=2.27x_0=2.27 , η(2.27)=0.457\eta(2.27)=0.457 Context question: A8 The band gap of pure silicon is Eg=1.11eVE_{\mathrm{g}}=1.11 \mathrm{eV}. Calculate the efficiency, ηSi\eta_{\mathrm{Si}}, of a silicon solar cell using this value. Context answer: 0.457 Extra Supplementary Reading Materials: In the late nineteenth century, Kelvin and Helmholtz (KH)(\mathrm{KH}) proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, M⊙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 potential energy through this slow contraction.

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