[
 {
  "problem_text": "Given that the work function for sodium metal is $2.28 \\mathrm{eV}$, what is the threshold frequency $v_0$ for sodium?",
  "answer_latex": " 5.51",
  "answer_number": "5.51",
  "unit": "$10^{14}\\mathrm{~Hz}$",
  "source": "chemmc",
  "problemid": " 1.1_3",
  "comment": " ",
  "solution": "\nWe must first convert $\\phi$ from electron volts to joules.\r\n$$\r\n\\begin{aligned}\r\n\\phi & =2.28 \\mathrm{eV}=(2.28 \\mathrm{eV})\\left(1.602 \\times 10^{-19} \\mathrm{~J} \\cdot \\mathrm{eV}^{-1}\\right) \\\\\r\n& =3.65 \\times 10^{-19} \\mathrm{~J}\r\n\\end{aligned}\r\n$$\r\nUsing Equation $h v_0=\\phi$, we have\r\n$$\r\nv_0=\\frac{3.65 \\times 10^{-19} \\mathrm{~J}}{6.626 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{s}}=5.51 \\times 10^{14} \\mathrm{~Hz}$$\n"
 },
 {
  "problem_text": "Calculate the de Broglie wavelength of an electron traveling at $1.00 \\%$ of the speed of light.",
  "answer_latex": " 243",
  "answer_number": "243",
  "unit": " $\\mathrm{pm}$",
  "source": "chemmc",
  "problemid": " 1.1_11",
  "comment": " ",
  "solution": "The mass of an electron is $9.109 \\times 10^{-31} \\mathrm{~kg}$. One percent of the speed of light is\r\n$$\r\nv=(0.0100)\\left(2.998 \\times 10^8 \\mathrm{~m} \\cdot \\mathrm{s}^{-1}\\right)=2.998 \\times 10^6 \\mathrm{~m} \\cdot \\mathrm{s}^{-1}\r\n$$\r\nThe momentum of the electron is given by\r\n$$\r\n\\begin{aligned}\r\np=m_{\\mathrm{e}} v & =\\left(9.109 \\times 10^{-31} \\mathrm{~kg}\\right)\\left(2.998 \\times 10^6 \\mathrm{~m} \\cdot \\mathrm{s}^{-1}\\right) \\\\\r\n& =2.73 \\times 10^{-24} \\mathrm{~kg} \\cdot \\mathrm{m} \\cdot \\mathrm{s}^{-1}\r\n\\end{aligned}\r\n$$\r\nThe de Broglie wavelength of this electron is\r\n$$\r\n\\begin{aligned}\r\n\\lambda=\\frac{h}{p} & =\\frac{6.626 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{s}}{2.73 \\times 10^{-24} \\mathrm{~kg} \\cdot \\mathrm{m} \\cdot \\mathrm{s}^{-1}}=2.43 \\times 10^{-10} \\mathrm{~m} \\\\\r\n& =243 \\mathrm{pm}\r\n\\end{aligned}\r\n$$\r\nThis wavelength is of atomic dimensions.\r\n"
 },
 {
  "problem_text": "Show that $u(\\theta, \\phi)=Y_1^1(\\theta, \\phi)$ given in Example $$\n\\begin{aligned}\n&Y_1^1(\\theta, \\phi)=-\\left(\\frac{3}{8 \\pi}\\right)^{1 / 2} e^{i \\phi} \\sin \\theta\\\\\n&Y_1^{-1}(\\theta, \\phi)=\\left(\\frac{3}{8 \\pi}\\right)^{1 / 2} e^{-i \\phi} \\sin \\theta\n\\end{aligned}\n$$ satisfies the equation $\\nabla^2 u=\\frac{c}{r^2} u$, where $c$ is a constant. What is the value of $c$ ?",
  "answer_latex": "-2 ",
  "answer_number": "-2",
  "unit": " ",
  "source": "chemmc",
  "problemid": " E.E_4",
  "comment": " ",
  "solution": "Because $u(\\theta, \\phi)$ is independent of $r$, we start with\r\n$$\r\n\\nabla^2 u=\\frac{1}{r^2 \\sin \\theta} \\frac{\\partial}{\\partial \\theta}\\left(\\sin \\theta \\frac{\\partial u}{\\partial \\theta}\\right)+\\frac{1}{r^2 \\sin ^2 \\theta} \\frac{\\partial^2 u}{\\partial \\phi^2}\r\n$$\r\nSubstituting\r\n$$\r\nu(\\theta, \\phi)=-\\left(\\frac{3}{8 \\pi}\\right)^{1 / 2} e^{i \\phi} \\sin \\theta\r\n$$\r\ninto $\\nabla^2 u$ gives\r\n$$\r\n\\begin{aligned}\r\n\\nabla^2 u & =-\\left(\\frac{3}{8 \\pi}\\right)^{1 / 2}\\left[\\frac{e^{i \\phi}}{r^2 \\sin \\theta}\\left(\\cos ^2 \\theta-\\sin ^2 \\theta\\right)-\\frac{\\sin \\theta}{r^2 \\sin ^2 \\theta} e^{i \\phi}\\right] \\\\\r\n& =-\\left(\\frac{3}{8 \\pi}\\right)^{1 / 2} \\frac{e^{i \\phi}}{r^2}\\left(\\frac{1-2 \\sin ^2 \\theta}{\\sin \\theta}-\\frac{1}{\\sin \\theta}\\right) \\\\\r\n& =2\\left(\\frac{3}{8 \\pi}\\right)^{1 / 2} \\frac{e^{i \\phi} \\sin \\theta}{r^2}\r\n\\end{aligned}\r\n$$\r\nor $c=-2$."
 },
 {
  "problem_text": "The wave function $\\Psi_2(1,2)$ given by Equation 9.39 is not normalized as it stands. Determine the normalization constant of $\\Psi_2(1,2)$ given that the \"1s\" parts are normalized.",
  "answer_latex": " $1 / \\sqrt{2}$",
  "answer_number": "0.70710678",
  "unit": " ",
  "source": "chemmc",
  "problemid": "9.9_4 ",
  "comment": " ",
  "solution": "We want to find the constant $c$ such that\r\n$$\r\nI=c^2\\left\\langle\\Psi_2(1,2) \\mid \\Psi_2(1,2)\\right\\rangle=1\r\n$$\r\nFirst notice that $\\Psi_2(1,2)$ can be factored into the product of a spatial part and a spin part:\r\n$$\r\n\\begin{aligned}\r\n\\Psi_2(1,2) & =1 s(1) 1 s(2)[\\alpha(1) \\beta(2)-\\alpha(2) \\beta(1)] \\\\\r\n& =1 s\\left(\\mathbf{r}_1\\right) 1 s\\left(\\mathbf{r}_2\\right)\\left[\\alpha\\left(\\sigma_1\\right) \\beta\\left(\\sigma_2\\right)-\\alpha\\left(\\sigma_2\\right) \\beta\\left(\\sigma_1\\right)\\right]\r\n\\end{aligned}\r\n$$\r\nThe normalization integral becomes the product of three integrals:\r\n$$\r\nI=c^2\\langle 1 s(1) \\mid 1 s(1)\\rangle\\langle 1 s(2) \\mid 1 s(2)\\rangle\\langle\\alpha(1) \\beta(1)-\\alpha(2) \\beta(1) \\mid \\alpha(1) \\beta(2)-\\alpha(2) \\beta(1)\\rangle\r\n$$\r\nThe spatial integrals are equal to 1 because we have taken the $1 s$ orbitals to be normalized. Now let's look at the spin integrals. When the two terms in the integrand of the spin integral are multiplied, we get four integrals. One of them is\r\n$$\r\n\\begin{aligned}\r\n\\iint \\alpha^*\\left(\\sigma_1\\right) \\beta^*\\left(\\sigma_2\\right) \\alpha\\left(\\sigma_1\\right) \\beta\\left(\\sigma_2\\right) d \\sigma_1 d \\sigma_2 & =\\langle\\alpha(1) \\beta(2) \\mid \\alpha(1) \\beta(2)\\rangle \\\\\r\n& =\\langle\\alpha(1) \\mid \\alpha(1)\\rangle\\langle\\beta(2) \\mid \\beta(2)\\rangle=1\r\n\\end{aligned}\r\n$$\r\nwhere once again we point out that integrating over $\\sigma_1$ and $\\sigma_2$ is purely symbolic; $\\sigma_1$ and $\\sigma_2$ are discrete variables. Another is\r\n$$\r\n\\langle\\alpha(1) \\beta(2) \\mid \\alpha(2) \\beta(1)\\rangle=\\langle\\alpha(1) \\mid \\beta(1)\\rangle\\langle\\beta(2) \\mid \\alpha(2)\\rangle=0\r\n$$\r\nThe other two are equal to 1 and 0 , and so\r\n$$\r\nI=c^2\\left\\langle\\Psi_2(1,2) \\mid \\Psi_2(1,2)\\right\\rangle=2 c^2=1\r\n$$\r\nor $c=1 / \\sqrt{2}$."
 },
 {
  "problem_text": "Find the bonding and antibonding H\u00fcckel molecular orbitals for ethene.",
  "answer_latex": " $1 / \\sqrt{2}$",
  "answer_number": "0.70710678",
  "unit": " ",
  "source": "chemmc",
  "problemid": "11.11_11 ",
  "comment": " ",
  "solution": "\nThe equations for $c_1$ and $c_2$ associated with Equation $$\n\\left|\\begin{array}{ll}\nH_{11}-E S_{11} & H_{12}-E S_{12} \\\\\nH_{12}-E S_{12} & H_{22}-E S_{22}\n\\end{array}\\right|=0\n$$ are\r\n$$\r\nc_1(\\alpha-E)+c_2 \\beta=0 \\quad \\text { and } \\quad c_1 \\beta+c_2(\\alpha-E)=0\r\n$$\r\nFor $E=\\alpha+\\beta$, either equation yields $c_1=c_2$. Thus,\r\n$$\r\n\\psi_{\\mathrm{b}}=c_1\\left(2 p_{z 1}+2 p_{z 2}\\right)\r\n$$\r\n\r\nThe value of $c_1$ can be found by requiring that the wave function be normalized. The normalization condition on $\\psi_\\pi$ gives $c_1^2(1+2 S+1)=1$. Using the H\u00fcckel assumption that $S=0$, we find that $c_1=1 / \\sqrt{2}$.\r\n\r\nSubstituting $E=\\alpha-\\beta$ into either of the equations for $c_1$ and $c_2$ yields $c_1=-c_2$, or\r\n$$\r\n\\psi_{\\mathrm{a}}=c_1\\left(2 p_{z 1}-2 p_{z 2}\\right)\r\n$$\r\nThe normalization condition gives $c^2(1-2 S+1)=1$, or $c_1=1 / \\sqrt{2}$.\n"
 },
 {
  "problem_text": "Using the explicit formulas for the Hermite polynomials given in Table 5.3 as below $$\n\\begin{array}{ll}\nH_0(\\xi)=1 & H_1(\\xi)=2 \\xi \\\\\nH_2(\\xi)=4 \\xi^2-2 & H_3(\\xi)=8 \\xi^3-12 \\xi \\\\\nH_4(\\xi)=16 \\xi^4-48 \\xi^2+12 & H_5(\\xi)=32 \\xi^5-160 \\xi^3+120 \\xi\n\\end{array}\n$$, show that a $0 \\rightarrow 1$ vibrational transition is allowed and that a $0 \\rightarrow 2$ transition is forbidden.",
  "answer_latex": "0 ",
  "answer_number": "0",
  "unit": " ",
  "source": "chemmc",
  "problemid": "5.5_12 ",
  "comment": " ",
  "solution": "Letting $\\xi=\\alpha^{1 / 2} x$ in Table 5.3 , we have\r\n$$\r\n\\begin{aligned}\r\n& \\psi_0(\\xi)=\\left(\\frac{\\alpha}{\\pi}\\right)^{1 / 4} e^{-\\xi^2 / 2} \\\\\r\n& \\psi_1(\\xi)=\\sqrt{2}\\left(\\frac{\\alpha}{\\pi}\\right)^{1 / 4} \\xi e^{-\\xi^2 / 2} \\\\\r\n& \\psi_2(\\xi)=\\frac{1}{\\sqrt{2}}\\left(\\frac{\\alpha}{\\pi}\\right)^{1 / 4}\\left(2 \\xi^2-1\\right) e^{-\\xi^2 / 2}\r\n\\end{aligned}\r\n$$\r\nThe dipole transition moment is given by the integral\r\n$$\r\nI_{0 \\rightarrow v} \\propto \\int_{-\\infty}^{\\infty} \\psi_v(\\xi) \\xi \\psi_0(\\xi) d \\xi\r\n$$\r\nThe transition is allowed if $I_{0 \\rightarrow v} \\neq 0$ and is forbidden if $I_{0 \\rightarrow v}=0$. For $v=1$, we have\r\n$$\r\nI_{0 \\rightarrow 1} \\propto\\left(\\frac{2 \\alpha}{\\pi}\\right)^{1 / 2} \\int_{-\\infty}^{\\infty} \\xi^2 e^{-\\xi^2} d \\xi \\neq 0\r\n$$\r\nbecause the integrand is everywhere positive. For $v=2$,\r\n$$\r\nI_{0 \\rightarrow 2} \\propto\\left(\\frac{\\alpha}{2 \\pi}\\right)^{1 / 2} \\int_{-\\infty}^{\\infty}\\left(2 \\xi^3-\\xi\\right) e^{-\\xi^2} d \\xi=0\r\n$$\r\nbecause the integrand is an odd function and the limits go from $-\\infty$ to $+\\infty$.\r\n"
 },
 {
  "problem_text": "To a good approximation, the microwave spectrum of $\\mathrm{H}^{35} \\mathrm{Cl}$ consists of a series of equally spaced lines, separated by $6.26 \\times 10^{11} \\mathrm{~Hz}$. Calculate the bond length of $\\mathrm{H}^{35} \\mathrm{Cl}$.",
  "answer_latex": " 129",
  "answer_number": "129",
  "unit": " $\\mathrm{pm}$",
  "source": "chemmc",
  "problemid": " 6.6_2",
  "comment": " ",
  "solution": "\nAccording to Equation $$\n\\begin{aligned}\n&v_{\\mathrm{obs}}=2 B(J+1) \\quad J=0,1,2, \\ldots\\\\\n&B=\\frac{h}{8 \\pi^2 I}\n\\end{aligned}\n$$, the spacing of the lines in the microwave spectrum of $\\mathrm{H}^{35} \\mathrm{Cl}$ is given by\r\n$$\r\n2 B=\\frac{h}{4 \\pi^2 I}\r\n$$\r\n\r\n$$\r\n\\frac{h}{4 \\pi^2 I}=6.26 \\times 10^{11} \\mathrm{~Hz}\r\n$$\r\nSolving this equation for $I$, we have\r\n$$\r\nI=\\frac{6.626 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{s}}{4 \\pi^2\\left(6.26 \\times 10^{11} \\mathrm{~s}^{-1}\\right)}=2.68 \\times 10^{-47} \\mathrm{~kg} \\cdot \\mathrm{m}^2\r\n$$\r\nThe reduced mass of $\\mathrm{H}^{35} \\mathrm{Cl}$ is\r\n$$\r\n\\mu=\\frac{(1.00 \\mathrm{amu})(35.0 \\mathrm{amu})}{36.0 \\mathrm{amu}}\\left(1.661 \\times 10^{-27} \\mathrm{~kg} \\cdot \\mathrm{amu}^{-1}\\right)=1.66 \\times 10^{-27} \\mathrm{~kg}\r\n$$\r\nUsing the fact that $I=\\mu l^2$, we obtain\r\n$$\r\nl=\\left(\\frac{2.68 \\times 10^{-47} \\mathrm{~kg} \\cdot \\mathrm{m}^2}{1.661 \\times 10^{-27} \\mathrm{~kg}}\\right)^{1 / 2}=1.29 \\times 10^{-10} \\mathrm{~m}=129 \\mathrm{pm}\r\n$$\n"
 },
 {
  "problem_text": "The unit of energy in atomic units is given by\r\n$$\r\n1 E_{\\mathrm{h}}=\\frac{m_{\\mathrm{e}} e^4}{16 \\pi^2 \\epsilon_0^2 \\hbar^2}\r\n$$\r\nExpress $1 E_{\\mathrm{h}}$ in electron volts $(\\mathrm{eV})$.",
  "answer_latex": "27.211 ",
  "answer_number": "27.211",
  "unit": " $\\mathrm{eV}$",
  "source": "chemmc",
  "problemid": "9.9_1 ",
  "comment": " ",
  "solution": "\nTo find $1 E_{\\mathrm{h}}$ expressed in joules, we substitute the SI values of $m_{\\mathrm{e}}, e$, $4 \\pi \\epsilon_0$, and $\\hbar$ into the above equation. Using these values from Table $$$\n\\begin{array}{lll}\n\\hline \\text { Property } & \\text { Atomic unit } & \\text { SI equivalent } \\\\\n\\hline \\text { Mass } & \\text { Mass of an electron, } m_{\\mathrm{e}} & 9.1094 \\times 10^{-31} \\mathrm{~kg} \\\\\n\\text { Charge } & \\text { Charge on a proton, } e & 1.6022 \\times 10^{-19} \\mathrm{C} \\\\\n\\text { Angular momentum } & \\text { Planck constant divided by } 2 \\pi, \\hbar & 1.0546 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{s} \\\\\n\\text { Length } & \\text { Bohr radius, } a_0=\\frac{4 \\pi \\epsilon_0 \\hbar^2}{m_{\\mathrm{e}} e^2} & 5.2918 \\times 10^{-11} \\mathrm{~m} \\\\\n\\text { Energy } & \\frac{m_{\\mathrm{e}} e^4}{16 \\pi^2 \\epsilon_0^2 \\hbar^2}=\\frac{e^2}{4 \\pi \\epsilon_0 a_0}=E_{\\mathrm{h}} & 4.3597 \\times 10^{-18} \\mathrm{~J} \\\\\n\\text { Permittivity } & \\kappa_0=4 \\pi \\epsilon_0 & 1.1127 \\times 10^{-10} \\mathrm{C}^2 \\cdot \\mathrm{J}^{-1} \\cdot \\mathrm{m}^{-1} \\\\\n\\hline\n\\end{array}\n$$, we find\r\nAtomic and Molecular Calculations Are Expressed in Atomic Units\r\n$$\r\n\\begin{aligned}\r\n1 E_{\\mathrm{h}} & =\\frac{\\left(9.1094 \\times 10^{-31} \\mathrm{~kg}\\right)\\left(1.6022 \\times 10^{-19} \\mathrm{C}\\right)^4}{\\left(1.1127 \\times 10^{-10} \\mathrm{C}^2 \\cdot \\mathrm{J}^{-1} \\cdot \\mathrm{m}^{-1}\\right)^2\\left(1.0546 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{s}\\right)^2} \\\\\r\n& =4.3597 \\times 10^{-18} \\mathrm{~J}\r\n\\end{aligned}\r\n$$\r\nIf we multiply this result by the Avogadro constant, we obtain\r\n$$\r\n1 E_{\\mathrm{h}}=2625.5 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}\r\n$$\r\nTo express $1 E_{\\mathrm{h}}$ in wave numbers $\\left(\\mathrm{cm}^{-1}\\right)$, we use the fact that $1 E_{\\mathrm{h}}=4.3597 \\times 10^{-18} \\mathrm{~J}$ along with the equation\r\n$$\r\n\\begin{aligned}\r\n\\tilde{v} & =\\frac{1}{\\lambda}=\\frac{h v}{h c}=\\frac{E}{c h}=\\frac{4.3597 \\times 10^{-18} \\mathrm{~J}}{\\left(2.9979 \\times 10^8 \\mathrm{~m} \\cdot \\mathrm{s}^{-1}\\right)\\left(6.6261 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{s}\\right)} \\\\\r\n& =2.1947 \\times 10^7 \\mathrm{~m}^{-1}=2.1947 \\times 10^5 \\mathrm{~cm}^{-1}\r\n\\end{aligned}\r\n$$\r\nso that we can write\r\n$$\r\n1 E_{\\mathrm{h}}=2.1947 \\times 10^5 \\mathrm{~cm}^{-1}\r\n$$\r\nLast, to express $1 E_{\\mathrm{h}}$ in terms of electron volts, we use the conversion factor\r\n$$\r\n1 \\mathrm{eV}=1.6022 \\times 10^{-19} \\mathrm{~J}\r\n$$\r\nUsing the value of $1 E_{\\mathrm{h}}$ in joules obtained previously, we have\r\n$$\r\n\\begin{aligned}\r\n1 E_{\\mathrm{h}} & =\\left(4.3597 \\times 10^{-18} \\mathrm{~J}\\right)\\left(\\frac{1 \\mathrm{eV}}{1.6022 \\times 10^{-19} \\mathrm{~J}}\\right) \\\\\r\n& =27.211 \\mathrm{eV}\r\n\\end{aligned}\r\n$$\n"
 },
 {
  "problem_text": "Calculate the probability that a particle in a one-dimensional box of length $a$ is found between 0 and $a / 2$.",
  "answer_latex": "$\\frac{1}{2}$",
  "answer_number": "0.5",
  "unit": " ",
  "source": "chemmc",
  "problemid": "3.3_6 ",
  "comment": " ",
  "solution": "The probability that the particle will be found between 0 and $a / 2$ is\r\n$$\r\n\\operatorname{Prob}(0 \\leq x \\leq a / 2)=\\int_0^{a / 2} \\psi^*(x) \\psi(x) d x=\\frac{2}{a} \\int_0^{a / 2} \\sin ^2 \\frac{n \\pi x}{a} d x\r\n$$\r\nIf we let $n \\pi x / a$ be $z$, then we find\r\n\r\n$$\r\n\\begin{aligned}\r\n\\operatorname{Prob}(0 \\leq x \\leq a / 2) & =\\frac{2}{n \\pi} \\int_0^{n \\pi / 2} \\sin ^2 z d z=\\frac{2}{n \\pi}\\left|\\frac{z}{2}-\\frac{\\sin 2 z}{4}\\right|_0^{n \\pi / 2} \\\\\r\n& =\\frac{2}{n \\pi}\\left(\\frac{n \\pi}{4}-\\frac{\\sin n \\pi}{4}\\right)=\\frac{1}{2} \\quad \\text { (for all } n \\text { ) }\r\n\\end{aligned}\r\n$$\r\nThus, the probability that the particle lies in one-half of the interval $0 \\leq x \\leq a$ is $\\frac{1}{2}$."
 }
]