[
 {
  "problem_text": "A one-particle, one-dimensional system has $\\Psi=a^{-1 / 2} e^{-|x| / a}$ at $t=0$, where $a=1.0000 \\mathrm{~nm}$. At $t=0$, the particle's position is measured. Find the probability that the measured value is between $x=0$ and $x=2 \\mathrm{~nm}$.",
  "answer_latex": " 0.4908",
  "answer_number": "0.4908",
  "unit": " ",
  "source": "quan",
  "problemid": " 1.6.1_b",
  "comment": " ",
  "solution": "\n$$\r\n\\begin{aligned}\r\n\\operatorname{Pr}(0 \\leq x \\leq 2 \\mathrm{~nm}) & =\\int_0^{2 \\mathrm{~nm}}|\\Psi|^2 d x=a^{-1} \\int_0^{2 \\mathrm{~nm}} e^{-2 x / a} d x \\\\\r\n& =-\\left.\\frac{1}{2} e^{-2 x / a}\\right|_0 ^{2 \\mathrm{~nm}}=-\\frac{1}{2}\\left(e^{-4}-1\\right)=0.4908\r\n\\end{aligned}\r\n$$\n"
 },
 {
  "problem_text": "Calculate the ground-state energy of the hydrogen atom using SI units and convert the result to electronvolts.",
  "answer_latex": "-13.598 ",
  "answer_number": "-13.598 ",
  "unit": " $\\mathrm{eV}$",
  "source": "quan",
  "problemid": " 6.6.1",
  "comment": " ",
  "solution": "\nThe $\\mathrm{H}$ atom ground-state energy with $n=1$ and $Z=1$ is $E=-\\mu e^4 / 8 h^2 \\varepsilon_0^2$. Use of equation $$\n\\mu_{\\mathrm{H}}=\\frac{m_e m_p}{m_e+m_p}=\\frac{m_e}{1+m_e / m_p}=\\frac{m_e}{1+0.000544617}=0.9994557 m_e\n$$ for $\\mu$ gives\r\n$$\r\n\\begin{gathered}\r\nE=-\\frac{0.9994557\\left(9.109383 \\times 10^{-31} \\mathrm{~kg}\\right)\\left(1.6021766 \\times 10^{-19} \\mathrm{C}\\right)^4}{8\\left(6.626070 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}\\right)^2\\left(8.8541878 \\times 10^{-12} \\mathrm{C}^2 / \\mathrm{N}-\\mathrm{m}^2\\right)^2} \\frac{Z^2}{n^2} \\\\\r\nE=-\\left(2.178686 \\times 10^{-18} \\mathrm{~J}\\right)\\left(Z^2 / n^2\\right)\\left[(1 \\mathrm{eV}) /\\left(1.6021766 \\times 10^{-19} \\mathrm{~J}\\right)\\right]\r\n\\end{gathered}\r\n$$\r\n$$\r\nE=-(13.598 \\mathrm{eV})\\left(Z^2 / n^2\\right)=-13.598 \\mathrm{eV}\r\n$$\r\na number worth remembering. The minimum energy needed to ionize a ground-state hydrogen atom is $13.598 \\mathrm{eV}$.\n"
 },
 {
  "problem_text": "Find the probability that the electron in the ground-state $\\mathrm{H}$ atom is less than a distance $a$ from the nucleus.",
  "answer_latex": " 0.323",
  "answer_number": "0.323",
  "unit": " ",
  "source": "quan",
  "problemid": "6.6.3 ",
  "comment": " ",
  "solution": "\nWe want the probability that the radial coordinate lies between 0 and $a$. This is found by taking the infinitesimal probability of being between $r$ and $r+d r$ and summing it over the range from 0 to $a$. This sum of infinitesimal quantities is the definite integral\r\n$$\r\n\\begin{aligned}\r\n\\int_0^a R_{n l}^2 r^2 d r & =\\frac{4}{a^3} \\int_0^a e^{-2 r / a} r^2 d r=\\left.\\frac{4}{a^3} e^{-2 r / a}\\left(-\\frac{r^2 a}{2}-\\frac{2 r a^2}{4}-\\frac{2 a^3}{8}\\right)\\right|_0 ^a \\\\\r\n& =4\\left[e^{-2}(-5 / 4)-(-1 / 4)\\right]=0.323\r\n\\end{aligned}\r\n$$\n"
 },
 {
  "problem_text": "A one-particle, one-dimensional system has $\\Psi=a^{-1 / 2} e^{-|x| / a}$ at $t=0$, where $a=1.0000 \\mathrm{~nm}$. At $t=0$, the particle's position is measured. Find the probability that the measured value lies between $x=1.5000 \\mathrm{~nm}$ and $x=1.5001 \\mathrm{~nm}$.",
  "answer_latex": " 4.979",
  "answer_number": "4.979",
  "unit": " $10^{-6}$",
  "source": "quan",
  "problemid": " 1.6.1_a",
  "comment": " ",
  "solution": "\nIn this tiny interval, $x$ changes by only $0.0001 \\mathrm{~nm}$, and $\\Psi$ goes from $e^{-1.5000} \\mathrm{~nm}^{-1 / 2}=0.22313 \\mathrm{~nm}^{-1 / 2}$ to $e^{-1.5001} \\mathrm{~nm}^{-1 / 2}=0.22311 \\mathrm{~nm}^{-1 / 2}$, so $\\Psi$ is nearly constant in this interval, and it is a very good approximation to consider this interval as infinitesimal. The desired probability is given as\r\n$$\r\n\\begin{aligned}\r\n|\\Psi|^2 d x=a^{-1} e^{-2|x| / a} d x & =(1 \\mathrm{~nm})^{-1} e^{-2(1.5 \\mathrm{~nm}) /(1 \\mathrm{~nm})}(0.0001 \\mathrm{~nm}) \\\\\r\n& =4.979 \\times 10^{-6}\r\n\\end{aligned}\r\n$$\n\n"
 },
 {
  "problem_text": "In this example, $2.50 \\mathrm{~mol}$ of an ideal gas with $C_{V, m}=12.47 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$ is expanded adiabatically against a constant external pressure of 1.00 bar. The initial temperature and pressure of the gas are $325 \\mathrm{~K}$ and $2.50 \\mathrm{bar}$, respectively. The final pressure is 1.25 bar. Calculate the final $ \\Delta U$.",
  "answer_latex": " -1.78",
  "answer_number": "-1.78",
  "unit": " $\\mathrm{~kJ}$",
  "source": "quan",
  "problemid": " 2.6",
  "comment": " ",
  "solution": "Because the process is adiabatic, $q=0$, and $\\Delta U=w$. Therefore,\r\n$$\r\n\\Delta U=n C_{\\mathrm{v}, m}\\left(T_f-T_i\\right)=-P_{e x t e r n a l}\\left(V_f-V_i\\right)\r\n$$\r\nUsing the ideal gas law,\r\n$$\r\n\\begin{aligned}\r\n& n C_{\\mathrm{v}, m}\\left(T_f-T_i\\right)=-n R P_{\\text {external }}\\left(\\frac{T_f}{P_f}-\\frac{T_i}{P_i}\\right) \\\\\r\n& T_f\\left(n C_{\\mathrm{v}, m}+\\frac{n R P_{\\text {external }}}{P_f}\\right)=T_i\\left(n C_{\\mathrm{v}, m}+\\frac{n R P_{\\text {external }}}{P_i}\\right) \\\\\r\n& T_f=T_i\\left(\\frac{C_{\\mathrm{v}, m}+\\frac{R P_{\\text {external }}}{P_i}}{C_{\\mathrm{v}, m}+\\frac{R P_{\\text {external }}}{P_f}}\\right) \\\\\r\n& =325 \\mathrm{~K} \\times\\left(\\frac{12.47 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}+\\frac{8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} \\times 1.00 \\mathrm{bar}}{2.50 \\mathrm{bar}}}{12.47 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}+\\frac{8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} \\times 1.00 \\mathrm{bar}}{1.25 \\mathrm{bar}}}\\right)=268 \\mathrm{~K} \\\\\r\n&\r\n\\end{aligned}\r\n$$\r\nWe calculate $\\Delta U=w$ from\r\n$$\r\n\\begin{aligned}\r\n\\Delta U & =n C_{V, m}\\left(T_f-T_i\\right)=2.5 \\mathrm{~mol} \\times 12.47 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} \\times(268 \\mathrm{~K}-325 \\mathrm{~K}) \\\\\r\n& =-1.78 \\mathrm{~kJ}\r\n\\end{aligned}\r\n$$"
 },
 {
  "problem_text": "Find $Y_l^m(\\theta, \\phi)$  for  $l=0$.",
  "answer_latex": "$\\frac{1}{\\sqrt{4 \\pi}}$",
  "answer_number": "0.28209479",
  "unit": " ",
  "source": "quan",
  "problemid": " 5.1",
  "comment": " ",
  "solution": "\nFor $l=0$, Eq. $$\nm=-l,-l+1,-l+2, \\ldots,-1,0,1, \\ldots, l-2, l-1, l\n$$ gives $m=0$, and $$\nS_{l, m}(\\theta)=\\sin ^{|m|} \\theta \\sum_{\\substack{j=1,3, \\ldots \\ \\text { or } j=0,2, \\ldots}}^{l-|m|} a_j \\cos ^j \\theta\n$$ becomes\r\n$$\r\nS_{0,0}(\\theta)=a_0\r\n$$\r\nThe normalization condition $$\n\\int_0^{\\infty}|R|^2 r^2 d r=1, \\quad \\int_0^\\pi|S|^2 \\sin \\theta d \\theta=1, \\quad \\int_0^{2 \\pi}|T|^2 d \\phi=1\n$$ gives$$\\begin{aligned}\r\n\\int_0^\\pi\\left|a_0\\right|^2 \\sin \\theta d \\theta=1 & =2\\left|a_0\\right|^2 \\\\\r\n\\left|a_0\\right| & =2^{-1 / 2}\r\n\\end{aligned}\r\n$$\r\nEquation $$\nY_l^m(\\theta, \\phi)=S_{l, m}(\\theta) T(\\phi)=\\frac{1}{\\sqrt{2 \\pi}} S_{l, m}(\\theta) e^{i m \\phi}\n$$ gives$$\r\nY_0^0(\\theta, \\phi)=\\frac{1}{\\sqrt{4 \\pi}}\r\n$$\n\n"
 },
 {
  "problem_text": "The lowest-frequency pure-rotational absorption line of ${ }^{12} \\mathrm{C}^{32} \\mathrm{~S}$ occurs at $48991.0 \\mathrm{MHz}$. Find the bond distance in ${ }^{12} \\mathrm{C}^{32} \\mathrm{~S}$.",
  "answer_latex": " 1.5377",
  "answer_number": "1.5377",
  "unit": " $10^{-10} \\mathrm{~m}$",
  "source": "quan",
  "problemid": "6.4 ",
  "comment": " ",
  "solution": "\nThe lowest-frequency rotational absorption is the $J=0 \\rightarrow 1$ line. $$h \\nu=E_{\\mathrm{upper}}-E_{\\mathrm{lower}}=\\frac{1(2) \\hbar^2}{2 \\mu d^2}-\\frac{0(1) \\hbar^2}{2 \\mu d^2}\r\n$$\r\nwhich gives $d=\\left(h / 4 \\pi^2 \\nu \\mu\\right)^{1 / 2}$. $$\r\n\\mu=\\frac{m_1 m_2}{m_1+m_2}=\\frac{12(31.97207)}{(12+31.97207)} \\frac{1}{6.02214 \\times 10^{23}} \\mathrm{~g}=1.44885 \\times 10^{-23} \\mathrm{~g}\r\n$$\r\nThe SI unit of mass is the kilogram, and\r\n$$\r\n\\begin{aligned}\r\nd=\\frac{1}{2 \\pi}\\left(\\frac{h}{\\nu_{0 \\rightarrow 1} \\mu}\\right)^{1 / 2} & =\\frac{1}{2 \\pi}\\left[\\frac{6.62607 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}}{\\left(48991.0 \\times 10^6 \\mathrm{~s}^{-1}\\right)\\left(1.44885 \\times 10^{-26} \\mathrm{~kg}\\right)}\\right]^{1 / 2} \\\\\r\n& =1.5377 \\times 10^{-10} \\mathrm{~m}\r\n\\end{aligned}\r\n$$\n"
 },
 {
  "problem_text": "The strongest infrared band of ${ }^{12} \\mathrm{C}^{16} \\mathrm{O}$ occurs at $\\widetilde{\\nu}=2143 \\mathrm{~cm}^{-1}$. Find the force constant of ${ }^{12} \\mathrm{C}^{16} \\mathrm{O}$. ",
  "answer_latex": " 1855",
  "answer_number": "1855",
  "unit": " $\\mathrm{~N} / \\mathrm{m}$",
  "source": "quan",
  "problemid": " 4.3",
  "comment": " ",
  "solution": "\nThe strongest infrared band corresponds to the $v=0 \\rightarrow 1$ transition. We approximate the molecular vibration as that of a harmonic oscillator. The equilibrium molecular vibrational frequency is approximately\r\n$$\r\n\\nu_e \\approx \\nu_{\\text {light }}=\\widetilde{\\nu} c=\\left(2143 \\mathrm{~cm}^{-1}\\right)\\left(2.9979 \\times 10^{10} \\mathrm{~cm} / \\mathrm{s}\\right)=6.424 \\times 10^{13} \\mathrm{~s}^{-1}\r\n$$\r\nTo relate $k$ to $\\nu_e$, we need the reduced mass $\\mu=m_1 m_2 /\\left(m_1+m_2\\right)$. One mole of ${ }^{12} \\mathrm{C}$ has a mass of $12 \\mathrm{~g}$ and contains Avogadro's number of atoms. Hence the mass of one atom of ${ }^{12} \\mathrm{C}$ is $(12 \\mathrm{~g}) /\\left(6.02214 \\times 10^{23}\\right)$. The reduced mass and force constant are\r\n$$\r\n\\begin{gathered}\r\n\\mu=\\frac{12(15.9949) \\mathrm{g}}{27.9949} \\frac{1}{6.02214 \\times 10^{23}}=1.1385 \\times 10^{-23} \\mathrm{~g} \\\\\r\nk=4 \\pi^2 \\nu_e^2 \\mu=4 \\pi^2\\left(6.424 \\times 10^{13} \\mathrm{~s}^{-1}\\right)^2\\left(1.1385 \\times 10^{-26} \\mathrm{~kg}\\right)=1855 \\mathrm{~N} / \\mathrm{m}\r\n\\end{gathered}\r\n$$\n"
 }
]