B.4 If , calculate the rms speed of the Be nuclei, , and hence estimate . · OlympiadBench #1073

Physics · modern-physics

Problem

B.4 If ΔErms=5.54×10−17 J\Delta E_{r m s}=5.54 \times 10^{-17} \mathrm{~J}, calculate the rms speed of the Be nuclei, VBe\mathrm{V}_{\mathrm{Be}}, and hence estimate TcT_{\mathrm{c}}. (Hint: ΔErms\Delta E_{r m s} depends on the rms value of the component of velocity along the line of sight). 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. Context question: 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}. Context answer: Ω=−35GM⊙2R⊙\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}} Context question: A.10 Estimate the maximum possible time, τKH\tau_{\mathrm{KH}} (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period. Context answer: 1.88×1071.88 \times 10^{7} Extra Supplementary Reading Materials: The τKH\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. B Neutrinos from the Sun : In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is: 41H→4He+2e++2ve4{ }^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 \mathrm{e}^{+}+2 v_{\mathrm{e}} The "electron neutrinos", vev_{\mathrm{e}}, produced in this reaction may be taken to be massless. They escape the Sun and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem. Context question: B1 Calculate the flux density, Φv\Phi_{v}, of the number of neutrinos arriving at the Earth, in units of m−2 s−1\mathrm{m}^{-2} \mathrm{~s}^{-1}. The energy released in the above reaction is ΔE=4.0×10−12 J\Delta E=4.0 \times 10^{-12} \mathrm{~J}. Assume that the energy radiated by the Sun is entirely due to this reaction. Context answer: 6.8×10146.8 \times 10^{14} Extra Supplementary Reading Materials: Travelling from the core of the Sun to the Earth, some of the electron neutrinos, vev_{e}, are converted to other types of neutrinos, vxv_{\mathrm{x}}. The efficiency of the detector for detecting vxv_{\mathrm{x}} is 1/61 / 6 of its efficiency for detecting vev_{\mathrm{e}}. If there is no neutrino conversion, we expect to detect an average of N1N_{1} neutrinos in a year. However, due to the conversion, an average of N2N_{2} neutrinos ( vev_{\mathrm{e}} and vxv_{\mathrm{x}} combined) are actually detected per year. Context question: B2 In terms of N1N_{1} and N2N_{2}, calculate what fraction, ff, of vev_{\mathrm{e}} is converted to vxv_{\mathrm{x}}. Context answer: f=65(1−N2N1)f=\frac{6}{5}(1-\frac{N_{2}}{N_{1}}) Extra Supplementary Reading Materials: 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, nn ), this radiation, called Cherenkov radiation, is emitted in the shape of a cone. Context question: B.3 Assume that an electron knocked out by a neutrino loses energy at a constant rate of α\alpha per unit time, while it travels through water. If this electron emits Cherenkov radiation for a time, Δt\Delta t, determine the energy imparted to this electron ( Eimparted E_{\text {imparted }} ) by the neutrino, in terms of α,Δt,n,me\alpha, \Delta t, n, m_{\mathrm{e}} and cc. (Assume the electron to be at rest before its interaction with the neutrino.) Context answer: Eimparted =αΔt+(nn2−1−1)mec2E_{\text {imparted }}=\alpha \Delta t+(\frac{n}{\sqrt{n^{2}-1}}-1) m_{\mathrm{e}} c^{2} Extra Supplementary Reading Materials: The fusion of H\mathrm{H} into He inside the Sun takes place in several steps. Nucleus of 7Be{ }^{7} \mathrm{Be} (rest mass, mBem_{\mathrm{Be}} ) is produced in one of these intermediate steps. Subsequently, it can absorb an electron, producing a 7Li{ }^{7} \mathrm{Li} nucleus (rest mass, mLi<mBem_{\mathrm{Li}}<m_{\mathrm{Be}} ) and emitting a vev_{\mathrm{e}}. The corresponding nuclear reaction is: 7Be+e−→7Li+ve{ }^{7} \mathrm{Be}+\mathrm{e}^{-} \rightarrow{ }^{7} \mathrm{Li}+v_{\mathrm{e}} When a Be nucleus ( mBe=11.65×10−27 kgm_{\mathrm{Be}}=11.65 \times 10^{-27} \mathrm{~kg} ) is at rest and absorbs an electron also at rest, the emitted neutrino has energy Ev=1.44×10−13 JE_{v}=1.44 \times 10^{-13} \mathrm{~J}. However, the Be nuclei are in random thermal motion due to the temperature TcT_{\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 (rms) value ΔErms\Delta E_{r m s}.

Browse the problem library