| [{"problem_text": "Calculating the maximum wavelength capable of photoejection\r\nA photon of radiation of wavelength $305 \\mathrm{~nm}$ ejects an electron from a metal with a kinetic energy of $1.77 \\mathrm{eV}$. Calculate the maximum wavelength of radiation capable of ejecting an electron from the metal.", "answer_latex": " 540", "answer_number": "540", "unit": "$\\mathrm{~nm}$ ", "source": "matter", "problemid": " 4.1", "comment": " with solution", "solution": "From the expression for the work function $\\Phi=h \\nu-E_{\\mathrm{k}}$ the minimum frequency for photoejection is\r\n$$\r\n\\nu_{\\min }=\\frac{\\Phi}{h}=\\frac{h v-E_{\\mathrm{k}}}{h} \\stackrel{\\nu=c \\mid \\lambda}{=} \\frac{c}{\\lambda}-\\frac{E_{\\mathrm{k}}}{h}\r\n$$\r\nThe maximum wavelength is therefore\r\n$$\r\n\\lambda_{\\max }=\\frac{c}{v_{\\min }}=\\frac{c}{c / \\lambda-E_{\\mathrm{k}} / h}=\\frac{1}{1 / \\lambda-E_{\\mathrm{k}} / h c}\r\n$$\r\nNow we substitute the data. The kinetic energy of the electron is\r\n$$\r\n\\begin{aligned}\r\n& E_k=1.77 \\mathrm{eV} \\times\\left(1.602 \\times 10^{-19} \\mathrm{JeV}^{-1}\\right)=2.83 \\ldots \\times 10^{-19} \\mathrm{~J} \\\\\r\n& \\frac{E_k}{h c}=\\frac{2.83 \\ldots \\times 10^{-19} \\mathrm{~J}}{\\left(6.626 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}\\right) \\times\\left(2.998 \\times 10^8 \\mathrm{~m} \\mathrm{~s}^{-1}\\right)}=1.42 \\ldots \\times 10^6 \\mathrm{~m}^{-1}\r\n\\end{aligned}\r\n$$\r\nTherefore, with\r\n$$\r\n\\begin{aligned}\r\n& 1 / \\lambda=1 / 305 \\mathrm{~nm}=3.27 \\ldots \\times 10^6 \\mathrm{~m}^{-1}, \\\\\r\n& \\lambda_{\\max }=\\frac{1}{\\left(3.27 \\ldots \\times 10^6 \\mathrm{~m}^{-1}\\right)-\\left(1.42 \\ldots \\times 10^6 \\mathrm{~m}^{-1}\\right)}=5.40 \\times 10^{-7} \\mathrm{~m}\r\n\\end{aligned}\r\n$$\r\nor $540 \\mathrm{~nm}$.\r\n"}, {"problem_text": "Estimate the molar volume of $\\mathrm{CO}_2$ at $500 \\mathrm{~K}$ and 100 atm by treating it as a van der Waals gas.", "answer_latex": " 0.366", "answer_number": "0.366", "unit": "$\\mathrm{dm}^3 \\mathrm{~mol}^{-1}$", "source": "matter", "problemid": " 36.1", "comment": " with solution", "solution": "According to Table 36.3, $a=3.610 \\mathrm{dm}^6$ atm $\\mathrm{mol}^{-2}$ and $b=4.29 \\times 10^{-2} \\mathrm{dm}^3 \\mathrm{~mol}^{-1}$. Under the stated conditions, $R T / p=0.410 \\mathrm{dm}^3 \\mathrm{~mol}^{-1}$. The coefficients in the equation for $V_{\\mathrm{m}}$ are therefore\r\n$$\r\n\\begin{aligned}\r\nb+R T / p & =0.453 \\mathrm{dm}^3 \\mathrm{~mol}^{-1} \\\\\r\na / p & =3.61 \\times 10^{-2}\\left(\\mathrm{dm}^3 \\mathrm{~mol}^{-1}\\right)^2 \\\\\r\na b / p & =1.55 \\times 10^{-3}\\left(\\mathrm{dm}^3 \\mathrm{~mol}^{-1}\\right)^3\r\n\\end{aligned}\r\n$$\r\nTherefore, on writing $x=V_{\\mathrm{m}} /\\left(\\mathrm{dm}^3 \\mathrm{~mol}^{-1}\\right)$, the equation to solve is\r\n$$\r\nx^3-0.453 x^2+\\left(3.61 \\times 10^{-2}\\right) x-\\left(1.55 \\times 10^{-3}\\right)=0\r\n$$\r\nThe acceptable root is $x=0.366$, which implies that $V_{\\mathrm{m}}=0.366$ $\\mathrm{dm}^3 \\mathrm{~mol}^{-1}$. The molar volume of a perfect gas under these conditions is $0.410 \\mathrm{dm}^3 \\mathrm{~mol}^{-1}$.\r\n"}, {"problem_text": "The single electron in a certain excited state of a hydrogenic $\\mathrm{He}^{+}$ion $(Z=2)$ is described by the wavefunction $R_{3,2}(r) \\times$ $Y_{2,-1}(\\theta, \\phi)$. What is the energy of its electron?", "answer_latex": " -6.04697", "answer_number": " -6.04697", "unit": "$ \\mathrm{eV}$ ", "source": "matter", "problemid": " 17.1", "comment": " with solution", "solution": "Replacing $\\mu$ by $m_{\\mathrm{e}}$ and using $\\hbar=h / 2 \\pi$, we can write the expression for the energy (eqn 17.7) as\r\n$$\r\nE_n=-\\frac{Z^2 m_e e^4}{8 \\varepsilon_0^2 h^2 n^2}=-\\frac{Z^2 h c \\tilde{R}_{\\infty}}{n^2}\r\n$$\r\nwith\r\n$$\r\n\\begin{aligned}\r\n& \\times \\underbrace{2.997926 \\times 10^{10} \\mathrm{~cm} \\mathrm{~s}^{-1}}_c \\\\\r\n& =109737 \\mathrm{~cm}^{-1} \\\\\r\n&\r\n\\end{aligned}\r\n$$\r\nand\r\n$$\r\n\\begin{aligned}\r\nh c \\tilde{R}_{\\infty}= & \\left(6.62608 \\times 10^{-34} \\mathrm{Js}\\right) \\times\\left(2.997926 \\times 10^{10} \\mathrm{~cm} \\mathrm{~s}^{-1}\\right) \\\\\r\n& \\times\\left(109737 \\mathrm{~cm}^{-1}\\right) \\\\\r\n= & 2.17987 \\times 10^{-18} \\mathrm{~J}\r\n\\end{aligned}\r\n$$\r\nTherefore, for $n=3$, the energy is\r\n$$\r\n\\begin{aligned}\r\n& E_3=-\\frac{\\overbrace{4}^{Z^2} \\times \\overbrace{2.17987 \\times 10^{-18} \\mathrm{~J}}^{h c \\tilde{R}_{\\infty}}}{\\underset{\\tilde{n}^2}{9}} \\\\\r\n& =-9.68831 \\times 10^{-19} \\mathrm{~J} \\\\\r\n&\r\n\\end{aligned}\r\n$$\r\nor $-0.968831 \\mathrm{aJ}$ (a, for atto, is the prefix that denotes $10^{-18}$ ). In some applications it is useful to express the energy in electronvolts $\\left(1 \\mathrm{eV}=1.602176 \\times 10^{-19} \\mathrm{~J}\\right)$; in this case, $E_3=-6.04697 \\mathrm{eV}$"}, {"problem_text": "Calculate the typical wavelength of neutrons after reaching thermal equilibrium with their surroundings at $373 \\mathrm{~K}$. For simplicity, assume that the particles are travelling in one dimension.", "answer_latex": " 226", "answer_number": "226", "unit": "$\\mathrm{pm}$", "source": "matter", "problemid": " 37.4", "comment": " with solution", "solution": "From the equipartition principle, we know that the mean translational kinetic energy of a neutron at a temperature $T$ travelling in the $x$-direction is $E_{\\mathrm{k}}=\\frac{1}{2} k T$. The kinetic energy is also equal to $p^2 / 2 m$, where $p$ is the momentum of the neutron and $m$ is its mass. Hence, $p=(m k T)^{1 / 2}$. It follows from the de Broglie relation $\\lambda=h / p$ that the neutron's wavelength is\r\n$$\r\n\\lambda=\\frac{h}{(m k T)^{1 / 2}}\r\n$$\r\nTherefore, at $373 \\mathrm{~K}$,\r\n$$\r\n\\begin{aligned}\r\n\\lambda & =\\frac{6.626 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}}{\\left\\{\\left(1.675 \\times 10^{-27} \\mathrm{~kg}\\right) \\times\\left(1.381 \\times 10^{-23} \\mathrm{~J} \\mathrm{~K}^{-1}\\right) \\times(373 \\mathrm{~K})\\right\\}^{1 / 2}} \\\\\r\n& =\\frac{6.626 \\times 10^{-34}}{\\left(1.675 \\times 10^{-27} \\times 1.381 \\times 10^{-23} \\times 373\\right)^{1 / 2}} \\frac{\\mathrm{kg} \\mathrm{m}^2 \\mathrm{~s}^{-1}}{\\left(\\mathrm{~kg}^2 \\mathrm{~m}^2 \\mathrm{~s}^{-2}\\right)^{1 / 2}} \\\\\r\n& =2.26 \\times 10^{-10} \\mathrm{~m}=226 \\mathrm{pm}\r\n\\end{aligned}\r\n$$\r\nwhere we have used $1 \\mathrm{~J}=1 \\mathrm{~kg} \\mathrm{~m}^2 \\mathrm{~s}^{-2}$."}, {"problem_text": "Using the perfect gas equation\r\nCalculate the pressure in kilopascals exerted by $1.25 \\mathrm{~g}$ of nitrogen gas in a flask of volume $250 \\mathrm{~cm}^3$ at $20^{\\circ} \\mathrm{C}$.", "answer_latex": " 435", "answer_number": "435", "unit": " $\\mathrm{kPa}$", "source": "matter", "problemid": " 1.1", "comment": " with equation", "solution": "The amount of $\\mathrm{N}_2$ molecules (of molar mass $28.02 \\mathrm{~g}$ $\\mathrm{mol}^{-1}$ ) present is\r\n$$\r\nn\\left(\\mathrm{~N}_2\\right)=\\frac{m}{M\\left(\\mathrm{~N}_2\\right)}=\\frac{1.25 \\mathrm{~g}}{28.02 \\mathrm{~g} \\mathrm{~mol}^{-1}}=\\frac{1.25}{28.02} \\mathrm{~mol}\r\n$$\r\nThe temperature of the sample is\r\n$$\r\nT / K=20+273.15, \\text { so } T=(20+273.15) \\mathrm{K}\r\n$$\r\nTherefore, after rewriting eqn 1.5 as $p=n R T / V$,\r\n$$\r\n\\begin{aligned}\r\np & =\\frac{\\overbrace{(1.25 / 28.02) \\mathrm{mol}}^n \\times \\overbrace{\\left(8.3145 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right)}^R \\times \\overbrace{(20+273.15) \\mathrm{K}}^{\\left(2.50 \\times 10^{-4}\\right) \\mathrm{m}^3}}{\\underbrace{R}_V} \\\\\r\n& =\\frac{(1.25 / 28.02) \\times(8.3145) \\times(20+273.15)}{2.50 \\times 10^{-4}} \\frac{\\mathrm{J}}{\\mathrm{m}^3} \\\\\r\n1 \\mathrm{Jm}^{-3} & =1 \\mathrm{~Pa} \\\\\r\n& \\stackrel{\\infty}{=} 4.35 \\times 10^5 \\mathrm{~Pa}=435 \\mathrm{kPa}\r\n\\end{aligned}\r\n$$\r\n"}, {"problem_text": "Determine the energies and degeneracies of the lowest four energy levels of an ${ }^1 \\mathrm{H}^{35} \\mathrm{Cl}$ molecule freely rotating in three dimensions. What is the frequency of the transition between the lowest two rotational levels? The moment of inertia of an ${ }^1 \\mathrm{H}^{35} \\mathrm{Cl}$ molecule is $2.6422 \\times 10^{-47} \\mathrm{~kg} \\mathrm{~m}^2$.\r\n", "answer_latex": " 635.7", "answer_number": "635.7", "unit": " $\\mathrm{GHz}$", "source": "matter", "problemid": " 14.1", "comment": " with solution", "solution": "First, note that\r\n$$\r\n\\frac{\\hbar^2}{2 I}=\\frac{\\left(1.055 \\times 10^{-34} \\mathrm{Js}^2\\right.}{2 \\times\\left(2.6422 \\times 10^{-47} \\mathrm{~kg} \\mathrm{~m}^2\\right)}=2.106 \\ldots \\times 10^{-22} \\mathrm{~J}\r\n$$\r\nor $0.2106 \\ldots$ zJ. We now draw up the following table, where the molar energies are obtained by multiplying the individual energies by Avogadro's constant:\r\n\\begin{tabular}{llll}\r\n\\hline$J$ & $E / z J$ & $E /\\left(\\mathrm{J} \\mathrm{mol}^{-1}\\right)$ & Degeneracy \\\\\r\n\\hline 0 & 0 & 0 & 1 \\\\\r\n1 & 0.4212 & 253.6 & 3 \\\\\r\n2 & 1.264 & 760.9 & 5 \\\\\r\n3 & 2.527 & 1522 & 7 \\\\\r\n\\hline\r\n\\end{tabular}\r\n\r\nThe energy separation between the two lowest rotational energy levels $\\left(J=0\\right.$ and 1 ) is $4.212 \\times 10^{-22} \\mathrm{~J}$, which corresponds to a photon frequency of\r\n$$\r\n\\nu=\\frac{\\Delta E}{h}=\\frac{4.212 \\times 10^{-22} \\mathrm{~J}}{6.626 \\times 10^{-34} \\mathrm{Js}}=6.357 \\times 10^{11} \\mathrm{~s}^{-1}=635.7 \\mathrm{GHz}\r\n$$"}, {"problem_text": "Using the Planck distribution\r\nCompare the energy output of a black-body radiator (such as an incandescent lamp) at two different wavelengths by calculating the ratio of the energy output at $450 \\mathrm{~nm}$ (blue light) to that at $700 \\mathrm{~nm}$ (red light) at $298 \\mathrm{~K}$.\r\n", "answer_latex": " 2.10", "answer_number": "2.10", "unit": " $10^{-16}$", "source": "matter", "problemid": " 4.1", "comment": " with solution", "solution": "At a temperature $T$, the ratio of the spectral density of states at a wavelength $\\lambda_1$ to that at $\\lambda_2$ is given by\r\n$$\r\n\\frac{\\rho\\left(\\lambda_1, T\\right)}{\\rho\\left(\\lambda_2, T\\right)}=\\left(\\frac{\\lambda_2}{\\lambda_1}\\right)^5 \\times \\frac{\\left(\\mathrm{e}^{h c / \\lambda_2 k T}-1\\right)}{\\left(\\mathrm{e}^{h c / \\lambda_1 k T}-1\\right)}\r\n$$\r\nInsert the data and evaluate this ratio.\r\nAnswer With $\\lambda_1=450 \\mathrm{~nm}$ and $\\lambda_2=700 \\mathrm{~nm}$,\r\n$$\r\n\\begin{aligned}\r\n\\frac{h c}{\\lambda_1 k T} & =\\frac{\\left(6.626 \\times 10^{-34} \\mathrm{Js}\\right) \\times\\left(2.998 \\times 10^8 \\mathrm{~m} \\mathrm{~s}^{-1}\\right)}{\\left(450 \\times 10^{-9} \\mathrm{~m}\\right) \\times\\left(1.381 \\times 10^{-23} \\mathrm{~J} \\mathrm{~K}^{-1}\\right) \\times(298 \\mathrm{~K})}=107.2 \\ldots \\\\\r\n\\frac{h c}{\\lambda_2 k T} & =\\frac{\\left(6.626 \\times 10^{-34} \\mathrm{Js}\\right) \\times\\left(2.998 \\times 10^8 \\mathrm{~m} \\mathrm{~s}^{-1}\\right)}{\\left(700 \\times 10^{-9} \\mathrm{~m}\\right) \\times\\left(1.381 \\times 10^{-23} \\mathrm{JK}^{-1}\\right) \\times(298 \\mathrm{~K})}=68.9 \\ldots\r\n\\end{aligned}\r\n$$\r\nand therefore\r\n$$\r\n\\begin{aligned}\r\n& \\frac{\\rho(450 \\mathrm{~nm}, 298 \\mathrm{~K})}{\\rho(700 \\mathrm{~nm}, 298 \\mathrm{~K})}=\\left(\\frac{700 \\times 10^{-9} \\mathrm{~m}}{450 \\times 10^{-9} \\mathrm{~m}}\\right)^5 \\times \\frac{\\left(\\mathrm{e}^{68.9 \\cdots}-1\\right)}{\\left(\\mathrm{e}^{107.2 \\cdots}-1\\right)} \\\\\r\n& =9.11 \\times\\left(2.30 \\times 10^{-17}\\right)=2.10 \\times 10^{-16}\r\n\\end{aligned}\r\n$$"}, {"problem_text": "Lead has $T_{\\mathrm{c}}=7.19 \\mathrm{~K}$ and $\\mathcal{H}_{\\mathrm{c}}(0)=63.9 \\mathrm{kA} \\mathrm{m}^{-1}$. At what temperature does lead become superconducting in a magnetic field of $20 \\mathrm{kA} \\mathrm{m}^{-1}$ ?", "answer_latex": " 6.0", "answer_number": "6.0", "unit": " $\\mathrm{~K}$", "source": "matter", "problemid": " 39.2", "comment": " with solution", "solution": "Rearrangement of eqn 39.7 gives\r\n$$\r\nT=T_{\\mathrm{c}}\\left(1-\\frac{\\mathcal{H}_{\\mathrm{c}}(T)}{\\mathcal{H}_{\\mathrm{c}}(0)}\\right)^{1 / 2}\r\n$$\r\nand substitution of the data gives\r\n$$\r\nT=(7.19 \\mathrm{~K}) \\times\\left(1-\\frac{20 \\mathrm{kAm}^{-1}}{63.9 \\mathrm{kAm}^{-1}}\\right)^{1 / 2}=6.0 \\mathrm{~K}\r\n$$\r\nThat is, lead becomes superconducting at temperatures below $6.0 \\mathrm{~K}$.\r\n"}, {"problem_text": "When an electric discharge is passed through gaseous hydrogen, the $\\mathrm{H}_2$ molecules are dissociated and energetically excited $\\mathrm{H}$ atoms are produced. If the electron in an excited $\\mathrm{H}$ atom makes a transition from $n=2$ to $n=1$, calculate the wavenumber of the corresponding line in the emission spectrum.", "answer_latex": " 82258", "answer_number": "82258", "unit": " $\\mathrm{~cm}^{-1}$", "source": "matter", "problemid": " 17.2", "comment": " with solution", "solution": "The wavenumber of the photon emitted when an electron makes a transition from $n_2=2$ to $n_1=1$ is given by\r\n$$\r\n\\begin{aligned}\r\n\\tilde{\\boldsymbol{v}} & =-\\tilde{R}_{\\mathrm{H}}\\left(\\frac{1}{n_2^2}-\\frac{1}{n_1^2}\\right) \\\\\r\n& =-\\left(109677 \\mathrm{~cm}^{-1}\\right) \\times\\left(\\frac{1}{2^2}-\\frac{1}{1^2}\\right) \\\\\r\n& =82258 \\mathrm{~cm}^{-1}\r\n\\end{aligned}\r\n$$"}, {"problem_text": "Calculate the shielding constant for the proton in a free $\\mathrm{H}$ atom.", "answer_latex": " 1.775", "answer_number": "1.775", "unit": " $10^{-5}$", "source": "matter", "problemid": " 48.2", "comment": " with solution", "solution": "The wavefunction for a hydrogen 1 s orbital is\r\n$$\r\n\\psi=\\left(\\frac{1}{\\pi a_0^3}\\right)^{1 / 2} \\mathrm{e}^{-r / a_0}\r\n$$\r\nso, because $\\mathrm{d} \\tau=r^2 \\mathrm{~d} r \\sin \\theta \\mathrm{d} \\theta \\mathrm{d} \\phi$, the expectation value of $1 / r$ is written as\r\n$$\r\n\\begin{aligned}\r\n\\left\\langle\\frac{1}{r}\\right\\rangle & =\\int \\frac{\\psi^* \\psi}{r} \\mathrm{~d} \\tau=\\frac{1}{\\pi a_0^3} \\int_0^{2 \\pi} \\mathrm{d} \\phi \\int_0^\\pi \\sin \\theta \\mathrm{d} \\theta \\int_0^{\\infty} r \\mathrm{e}^{-2 r / a_0} \\mathrm{~d} r \\\\\r\n& =\\frac{4}{a_0^3} \\overbrace{\\int_0^{\\infty} r \\mathrm{e}^{-2 r / a_0} \\mathrm{~d} r}^{a_0^2 / 4 \\text { (Integral E.1) }}=\\frac{1}{a_0}\r\n\\end{aligned}\r\n$$\r\nwhere we used the integral listed in the Resource section. Therefore,\r\n$$\r\n\\begin{aligned}\r\n& =\\frac{\\left(1.602 \\times 10^{-19} \\mathrm{C}\\right)^2 \\times(4 \\pi \\times 10^{-7} \\overbrace{\\mathrm{J}}^{\\mathrm{Jg} \\mathrm{m}^2 \\mathrm{~s}^{-2}} \\mathrm{~s}^2 \\mathrm{C}^{-2} \\mathrm{~m}^{-1})}{12 \\pi \\times\\left(9.109 \\times 10^{-31} \\mathrm{~kg}\\right) \\times\\left(5.292 \\times 10^{-11} \\mathrm{~m}\\right)} \\\\\r\n& =1.775 \\times 10^{-5} \\\\\r\n&\r\n\\end{aligned}\r\n$$\r\n"}] |