physics_atoms
 Atoms Model of atom as per (J. J. Thomson) in 1898.The positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon. This model was called plum pudding model of the atom. Repulsion between alpha partcle and positively charged nucleusF=14𝜋𝜖0(2e)(Ze)r2$F=\frac{1}{4𝜋{𝜖}_{0}}\frac{\left(2e\right)\left(Ze\right)}{{r}^{2}}$ Electron Orbits for Hydrogen atomamv2r=14𝜋𝜖0e2r2Orbir radius=r=e24𝜋𝜖0mv2$\begin{array}{l}\frac{m{v}^{2}}{r}=\frac{1}{4𝜋{𝜖}_{0}}\frac{{e}^{2}}{{r}^{2}}\\ Orbir\phantom{\rule{0.22em}{0ex}}radius=r=\frac{{e}^{2}}{4𝜋{𝜖}_{0}m{v}^{2}}\end{array}$ Kinetic energy of the electron (Hydrogen Atom)K=12mv2=e28𝜋𝜖0r$K=\frac{1}{2}m{v}^{2}=\frac{{e}^{2}}{8𝜋{𝜖}_{0}r}$ Potential energy of the electron (Hydrogen Atom)U=- e24𝜋𝜖0r$U=-\phantom{\rule{0.22em}{0ex}}\frac{{e}^{2}}{4𝜋{𝜖}_{0}r}$ Total energy of the electron (Hydrogen Atom)E=K+U=- e28𝜋𝜖0r$E=K+U=-\phantom{\rule{0.22em}{0ex}}\frac{{e}^{2}}{8𝜋{𝜖}_{0}r}$ According to classical electromagnetic theory, an accelerating charged particle emits radiation in the form of electromagnetic waves. According to the classical electromagnetic theory, the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of revolution. Bohr’s postulate(i) Bohr’s first postulate : An electron in an atom could revolve in certain stable orbits without the emission of radiant energy, contrary to the predictions of electromagnetic theory. Each atom has certain definite stable states in which it can exist, and each possible state has definite total energy. These are called the stationary states of the atom. (ii) Bohr’s second postulate : This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h2 𝜋$\frac{h}{2\phantom{\rule{0.22em}{0ex}}𝜋}$ where h is the Planck’s constant (= 6.6×10-34 J s$=\phantom{\rule{0.22em}{0ex}}6.6×{10}^{-34}\phantom{\rule{0.22em}{0ex}}J\phantom{\rule{0.22em}{0ex}}s$). (iii) Bohr’s third postulate : It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. As per Bohr's principal, Radius of nth${n}^{th}$ orbit (for hydrogen):rn=(n2m)(h2𝜋)24𝜋𝜖0e2 ${r}_{n}=\left(\frac{{n}^{2}}{m}\right)\left(\frac{h}{2𝜋}{\right)}^{2}\frac{4𝜋{𝜖}_{0}}{{e}^{2}}\phantom{\rule{0.44em}{0ex}}$ The energy of an atom is the least (largest negative value) when its electron is revolving in an orbit closest to the nucleus i.e., the one for which n = 1. The lowest state of the atom, called the ground state, is that of the lowest energy, with the electron revolving in the orbit of smallest. The highest energy state corresponds to n =∞$\infty$ and has an energy of 0 eV. Total Energy of the electron in nth orbit for hydrogen atomEn=- me48 n2 𝜖20 h2${E}_{n}=-\phantom{\rule{0.22em}{0ex}}\frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{n}^{2}\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{2}}$ En=- 2.18×10-18n2 J${E}_{n}=-\phantom{\rule{0.22em}{0ex}}\frac{2.18×{10}^{-18}}{{n}^{2}}\phantom{\rule{0.22em}{0ex}}J$ En=- 13.6n2 eV${E}_{n}=-\phantom{\rule{0.22em}{0ex}}\frac{13.6}{{n}^{2}}\phantom{\rule{0.22em}{0ex}}eV$
Atomic Spectra
 Emmision Line Spectrum: When an atomic gas or vapour is excited at low pressure, by passing an electric current through it, the emitted radiation has a spectrum which contains certain specific wave lengths only. A spectrum of this kind is termed as emission line spectrum and it consists of bright lines on a dark background. The emission spectrum is a range of frequencies of electromagnetic radiation emitted by an atom or molecule after passing from a high energy state (excited state) to a lower one or configuration with less energy).
Line Spectra of Hydrogen Atom
 When an atom makes a transition from the higher energy state with quantum number ni${n}_{i}$ to the lower energy state nf${n}_{f}$ the difference of energy is carried away by photon of frequency 𝜈if${𝜈}_{if}$ah𝜈if=Eni-Enf⟹h𝜈if=me48 𝜖20 h2(1n 2f-1n 2i) ⟹𝜈if=me48 𝜖20 h3(1n 2f-1n 2i) Rydberg Constant=R=me48 𝜖20 h3c $\begin{array}{l}h{𝜈}_{if}={{E}_{n}}_{i}-{{E}_{n}}_{f}\\ ⟹h{𝜈}_{if}=\frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{2}}\left(\frac{1}{{n}_{f}^{\phantom{\rule{0.22em}{0ex}}2}}-\frac{1}{{n}_{i}^{\phantom{\rule{0.22em}{0ex}}2}}\right)\phantom{\rule{0.22em}{0ex}}\\ \\ ⟹{𝜈}_{if}=\frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{3}}\left(\frac{1}{{n}_{f}^{\phantom{\rule{0.22em}{0ex}}2}}-\frac{1}{{n}_{i}^{\phantom{\rule{0.22em}{0ex}}2}}\right)\phantom{\rule{0.22em}{0ex}}\\ \\ Rydberg\phantom{\rule{0.22em}{0ex}}Constant=R=\frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{3}c}\phantom{\rule{0.22em}{0ex}}\\ \\ \end{array}$ Lyman Series : 𝜈=Rc(112-1n2)$𝜈=Rc\left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)$ where , n = 2, 3, 4, 5 . . Balmer Series : 𝜈=Rc(122-1n2)$𝜈=Rc\left(\frac{1}{{2}^{2}}-\frac{1}{{n}^{2}}\right)$ where , n = 3, 4, 5 . . Paschen Series : 𝜈=Rc(132-1n2)$𝜈=Rc\left(\frac{1}{{3}^{2}}-\frac{1}{{n}^{2}}\right)$ where , n = 4, 5 . . Brackett Series : 𝜈=Rc(142-1n2)$𝜈=Rc\left(\frac{1}{{4}^{2}}-\frac{1}{{n}^{2}}\right)$ where , n = 5, 6 . . Pfund Series : 𝜈=Rc(152-1n2)$𝜈=Rc\left(\frac{1}{{5}^{2}}-\frac{1}{{n}^{2}}\right)$ where , n = 6, 7, 8 . .
 Questions Q.12.1 (a) The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model. (b) In the ground state of Thomson’s model electrons are in stable equilibrium, while in Rutherford’s model electrons always experience a net force. (c) A classical atom based on Rutherford’s model is doomed to collapse. (d) An atom has a nearly continuous mass distribution in a Thomson’s model but has a highly non-uniform mass distribution in Rutherford’s model. (e) The positively charged part of the atom possesses most of the mass in both the models. Q.12.2 Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect? A.12.2Hydrogen is much lighter target compared to the alpha particle because mass of hydrogen is mush less than 𝛼-particle$𝛼-particle$. So alpha particle will not bounce back (return back), if solid hydrogen is used. So we cannot detrmine the size of the nucleus. Q.12.3What is the shortest wavelength present in the Paschen series of spectral lines? A.12.3Paschen Series : 𝜈=Rc(132-1n2)$𝜈=Rc\left(\frac{1}{{3}^{2}}-\frac{1}{{n}^{2}}\right)$ where , n = 4, 5 . . For shortest wavelength 𝜈 (frequency)$𝜈\phantom{\rule{0.22em}{0ex}}\left(frequency\right)$ must be maximum. this will happen when n=infinitya𝜈=1.097×107×c×(132-1∞2)=0.122×107 c ⟹𝜈=0.122×107 c⟹𝜆=c𝜈=10.122×107=8.196×10-7=8.196×10-9 ⟹𝜆=819.6 nm $\begin{array}{l}𝜈=1.097×{10}^{7}×c×\left(\frac{1}{{3}^{2}}-\frac{1}{{\infty }^{2}}\right)=0.122×{10}^{7}\phantom{\rule{0.22em}{0ex}}c\phantom{\rule{0.22em}{0ex}}\\ ⟹𝜈=0.122×{10}^{7}\phantom{\rule{0.22em}{0ex}}c\\ ⟹𝜆=\frac{c}{𝜈}=\frac{1}{0.122×{10}^{7}}=8.196×{10}^{-7}=8.196×{10}^{-9}\\ \\ ⟹𝜆=819.6\phantom{\rule{0.22em}{0ex}}nm\\ \end{array}$ Q.12.4A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?A.12.4The frequency of the emitted photon is given by𝛥E=h𝜈$𝛥E=h𝜈$𝛥E=2.3 eV=2.3×1.6×10-19 Joules $𝛥E=2.3\phantom{\rule{0.22em}{0ex}}eV=2.3×1.6×{10}^{-19}\phantom{\rule{0.22em}{0ex}}Joules\phantom{\rule{0.22em}{0ex}}$h=6.626×10-34 m2kg s-1 $h=6.626×{10}^{-34}\phantom{\rule{0.22em}{0ex}}{m}^{2}kg\phantom{\rule{0.22em}{0ex}}{s}^{-1}\phantom{\rule{0.22em}{0ex}}$𝜈=𝛥Eh= 2.3×1.6×10-166.626×10-34=5.554×1014 Hz $𝜈=\frac{𝛥E}{h}=\frac{\phantom{\rule{0.22em}{0ex}}2.3×1.6×{10}^{-16}}{6.626×{10}^{-34}}=5.554×{10}^{14}\phantom{\rule{0.44em}{0ex}}Hz\phantom{\rule{0.44em}{0ex}}$ Q.12.5The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?A.12.5For hydrogen atomPotential energy : U=- e24𝜋𝜖0r$U=-\phantom{\rule{0.22em}{0ex}}\frac{{e}^{2}}{4𝜋{𝜖}_{0}r}$ Kinetic energy K=e28𝜋𝜖0r$K=\frac{{e}^{2}}{8𝜋{𝜖}_{0}r}$ So, let Kinetic enegy = x then Poential energy = -2x ⟹ x+(-2x)=-13.6$⟹\phantom{\rule{0.22em}{0ex}}x+\left(-2x\right)=-13.6$⟹ -x=-13.6$⟹\phantom{\rule{0.22em}{0ex}}-x=-13.6$⟹ x=13.6$⟹\phantom{\rule{0.22em}{0ex}}x=13.6$⟹ -2x=-27.2$⟹\phantom{\rule{0.22em}{0ex}}-2x=-27.2$ So, Kinetic energy=13.6 eVPotential Energy= - 27.2 eV Q.12.6A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.A.12.6 1𝜆=R(112-142)=1.097×107(11-116)$\frac{1}{𝜆}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{4}^{2}}\right)=1.097×{10}^{7}\left(\frac{1}{1}-\frac{1}{16}\right)$1𝜆=1.097×107(1516)$\frac{1}{𝜆}=1.097×{10}^{7}\left(\frac{15}{16}\right)$𝜆=0.9723×10-7=97.23 nm$𝜆=0.9723×{10}^{-7}=97.23\phantom{\rule{0.22em}{0ex}}nm$frequency=𝜈=c𝜆=3×1080.9723×10-7$frequency=𝜈=\frac{c}{𝜆}=\frac{3×{10}^{8}}{0.9723×{10}^{-7}}$frequency=𝜈=3.1×1015 Hz$frequency=𝜈=3.1×{10}^{15}\phantom{\rule{0.22em}{0ex}}Hz$ Q.12.8The radius of the innermost electron orbit of a hydrogen atom is 5.3×10-11m$5.3×{10}^{-11}m$. What are the radii of the n = 2 and n =3 orbits?A.12.8As per Bohr's principal, Radius of nth${n}^{th}$ orbit:rn=(n2m)(h2𝜋)24𝜋𝜖0e2 ${r}_{n}=\left(\frac{{n}^{2}}{m}\right)\left(\frac{h}{2𝜋}{\right)}^{2}\frac{4𝜋{𝜖}_{0}}{{e}^{2}}\phantom{\rule{0.44em}{0ex}}$ So, rn∝n2$So,\phantom{\rule{0.22em}{0ex}}{r}_{n}\propto {n}^{2}$ ar1r2=n21n22 ⟹5.3×10-11r2=1222 ⟹r2=21.2×10-11=2.12×10-10m$\begin{array}{l}\frac{{r}_{1}}{{r}_{2}}=\frac{{n}_{1}^{2}}{{n}_{2}^{2}}\\ \\ ⟹\frac{5.3×{10}^{-11}}{{r}_{2}}=\frac{{1}^{2}}{{2}^{2}}\phantom{\rule{0.22em}{0ex}}\\ \\ ⟹{r}_{2}=21.2×{10}^{-11}=2.12×{10}^{-10}m\end{array}$ ar1r3=n21n23 ⟹5.3×10-11r3=1232 ⟹r3=47.7×10-11=4.77×10-10m$\begin{array}{l}\frac{{r}_{1}}{{r}_{3}}=\frac{{n}_{1}^{2}}{{n}_{3}^{2}}\\ \\ ⟹\frac{5.3×{10}^{-11}}{{r}_{3}}=\frac{{1}^{2}}{{3}^{2}}\phantom{\rule{0.22em}{0ex}}\\ \\ ⟹{r}_{3}=47.7×{10}^{-11}=4.77×{10}^{-10}m\end{array}$ Q.12.9A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?A.12.9 Total energy of the electron in the stationary states of the hydrogen :En=-2.18×10-18n2 Joules=-13.6n2 eV ${E}_{n}=-\frac{2.18×{10}^{-18}}{{n}^{2}}\phantom{\rule{0.22em}{0ex}}Joules=-\frac{13.6}{{n}^{2}}\phantom{\rule{0.22em}{0ex}}eV\phantom{\rule{0.22em}{0ex}}$ E1=-13.6 eV ${E}_{1}=-13.6\phantom{\rule{0.22em}{0ex}}eV\phantom{\rule{0.22em}{0ex}}$aEnergy of electron beam=12.5 eVSo, energy of lydrogen becomes=-13.6+12.5=-1.1 eV$\begin{array}{l}Energy\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}electron\phantom{\rule{0.22em}{0ex}}beam=12.5\phantom{\rule{0.22em}{0ex}}eV\\ So,\phantom{\rule{0.22em}{0ex}}energy\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}lydrogen\phantom{\rule{0.22em}{0ex}}becomes=-13.6+12.5=-1.1\phantom{\rule{0.22em}{0ex}}eV\end{array}$ En=-13.6n2 eV ${E}_{n}=-\frac{13.6}{{n}^{2}}\phantom{\rule{0.22em}{0ex}}eV\phantom{\rule{0.22em}{0ex}}$ E1=-13.6 eV${E}_{1}=-13.6\phantom{\rule{0.44em}{0ex}}eV$E2=-3.4 eV${E}_{2}=-3.4\phantom{\rule{0.44em}{0ex}}eV$E3=-1.5 eV${E}_{3}=-1.5\phantom{\rule{0.44em}{0ex}}eV$E4=-1.01 eV${E}_{4}=-1.01\phantom{\rule{0.22em}{0ex}}eV$ As, E4>-1.1${E}_{4}>-1.1$ so electron can jump from n=1 to n=3 While de-exiting electrom can move from n=3 to n=1 From 3 to 2 will be Balmer seriesa1𝜆=R(122-132)=1.097×107(14-19)⟹1𝜆=1.522×106m⟹𝜆=656.3 nm $\begin{array}{l}\frac{1}{𝜆}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right)=1.097×{10}^{7}\left(\frac{1}{4}-\frac{1}{9}\right)\\ ⟹\frac{1}{𝜆}=1.522×{10}^{6}m\\ ⟹𝜆=656.3\phantom{\rule{0.22em}{0ex}}nm\phantom{\rule{0.22em}{0ex}}\end{array}$ From 3 to 1 will be Lyman seriesa1𝜆=R(112-132)=1.097×107(11-19)⟹1𝜆=9.751×106m⟹𝜆=102.55 nm ≈103 nm$\begin{array}{l}\frac{1}{𝜆}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{3}^{2}}\right)=1.097×{10}^{7}\left(\frac{1}{1}-\frac{1}{9}\right)\\ ⟹\frac{1}{𝜆}=9.751×{10}^{6}m\\ ⟹𝜆=102.55\phantom{\rule{0.22em}{0ex}}nm\phantom{\rule{0.44em}{0ex}}\approx 103\phantom{\rule{0.22em}{0ex}}nm\end{array}$ From 2 to 1 will be Lyman seriesa1𝜆=R(112-122)=1.097×107(11-14)⟹1𝜆=8.2275×106m⟹𝜆=121.55 nm ≈122 nm$\begin{array}{l}\frac{1}{𝜆}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)=1.097×{10}^{7}\left(\frac{1}{1}-\frac{1}{4}\right)\\ ⟹\frac{1}{𝜆}=8.2275×{10}^{6}m\\ ⟹𝜆=121.55\phantom{\rule{0.22em}{0ex}}nm\phantom{\rule{0.22em}{0ex}}\approx 122\phantom{\rule{0.22em}{0ex}}nm\end{array}$ Q.12.13 Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit. A.12.13 𝜈if=me48 𝜖20 h3(1n 2f-1n 2i) ${𝜈}_{if}=\frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{3}}\left(\frac{1}{{n}_{f}^{\phantom{\rule{0.22em}{0ex}}2}}-\frac{1}{{n}_{i}^{\phantom{\rule{0.22em}{0ex}}2}}\right)\phantom{\rule{0.22em}{0ex}}$ when hydrogen atom de-excites from level n to level (n–1) then 𝜈=me48 𝜖20 h3(1(n-1) 2-1n 2)=me48 𝜖20 h3(2n-1n2(n-1) 2) ${𝜈}_{}=\frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{3}}\left(\frac{1}{\left(n-1{\right)}_{}^{\phantom{\rule{0.22em}{0ex}}2}}-\frac{1}{{n}_{}^{\phantom{\rule{0.22em}{0ex}}2}}\right)=\frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{3}}\left(\frac{2n-1}{{n}^{2}\left(n-1{\right)}_{}^{\phantom{\rule{0.22em}{0ex}}2}}\right)\phantom{\rule{0.22em}{0ex}}$ considering n is very large 𝜈≈me48 𝜖20 h3(2nn4) $𝜈\approx \frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{3}}\left(\frac{2n}{{n}^{4}}\right)\phantom{\rule{0.22em}{0ex}}$𝜈≈me48 𝜖20 h3(2n3) -------------- (i)$𝜈\approx \frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{3}}\left(\frac{2}{{n}^{3}}\right)\phantom{\rule{0.22em}{0ex}}--------------\phantom{\rule{0.22em}{0ex}}\left(i\right)$ Now mev2er=14𝜋𝜖0e2r2 $\frac{{m}_{e}{v}_{e}^{2}}{r}=\frac{1}{4𝜋{𝜖}_{0}}\frac{{e}^{2}}{{r}^{2}}\phantom{\rule{0.66em}{0ex}}$⟹ mev2e=14𝜋𝜖0e2r ---------- (ii) $⟹\phantom{\rule{0.22em}{0ex}}{m}_{e}{v}_{e}^{2}=\frac{1}{4𝜋{𝜖}_{0}}\frac{{e}^{2}}{r}\phantom{\rule{0.66em}{0ex}}----------\phantom{\rule{0.22em}{0ex}}\left(ii\right)\phantom{\rule{0.22em}{0ex}}$ me ve r=nh2𝜋${m}_{e\phantom{\rule{0.22em}{0ex}}}{v}_{e}\phantom{\rule{0.22em}{0ex}}r=\frac{nh}{2𝜋}$ ⟹ve =nh2𝜋 me r ---------- (iii) $⟹{v}_{e}\phantom{\rule{0.22em}{0ex}}=\frac{nh}{2𝜋\phantom{\rule{0.22em}{0ex}}{m}_{e}\phantom{\rule{0.22em}{0ex}}r\phantom{\rule{0.22em}{0ex}}}\phantom{\rule{2.86em}{0ex}}----------\phantom{\rule{0.22em}{0ex}}\left(iii\right)\phantom{\rule{0.22em}{0ex}}$ By (i) and (ii) me n2h24𝜋2 m2e r2 =14𝜋𝜖0e2r$\frac{\phantom{\rule{0.44em}{0ex}}{m}_{e}\phantom{\rule{0.22em}{0ex}}{n}^{2}{h}^{2}}{4{𝜋}^{2}\phantom{\rule{0.44em}{0ex}}{m}_{e}^{2}\phantom{\rule{0.22em}{0ex}}{r}^{2}\phantom{\rule{0.22em}{0ex}}}=\frac{1}{4𝜋{𝜖}_{0}}\frac{{e}^{2}}{r}$ ⟹ r=n2h2 𝜖0 𝜋 me e2$⟹\phantom{\rule{0.22em}{0ex}}r=\frac{{n}^{2}{h}^{2}\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}\phantom{\rule{0.44em}{0ex}}}{𝜋\phantom{\rule{0.22em}{0ex}}{m}_{e}^{}\phantom{\rule{0.22em}{0ex}}{e}^{2}}$ Frequency of revolution of electron=𝜈e$={𝜈}_{e}$ 𝜈e=𝜔2𝜋=ve2𝜋r=nh4𝜋2 me r2 =nh4𝜋2 me 𝜋2 m2e e4 n4h4 𝜖20 ${𝜈}_{e}=\frac{𝜔}{2𝜋}=\frac{{v}_{e}}{2𝜋r}=\frac{nh}{4{𝜋}^{2}\phantom{\rule{0.22em}{0ex}}{m}_{e}\phantom{\rule{0.22em}{0ex}}{r}^{2}\phantom{\rule{0.22em}{0ex}}}=\frac{nh}{4{𝜋}^{2}\phantom{\rule{0.22em}{0ex}}{m}_{e}\phantom{\rule{0.22em}{0ex}}}\frac{{𝜋}^{2}\phantom{\rule{0.22em}{0ex}}{m}_{e}^{2}\phantom{\rule{0.22em}{0ex}}{e}^{4}\phantom{\rule{0.44em}{0ex}}}{{n}^{4}{h}^{4}\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}}\phantom{\rule{0.22em}{0ex}}$ 𝜈e=me e44 h3 n3𝜖20 =me48 𝜖20 h3(2n3) ---------(iv) ${𝜈}_{e}=\frac{{m}_{e}^{}\phantom{\rule{0.22em}{0ex}}{e}^{4}}{4\phantom{\rule{0.22em}{0ex}}{h}^{3}\phantom{\rule{0.22em}{0ex}}{n}^{3}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}}=\frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{3}}\left(\frac{2}{{n}^{3}}\right)\phantom{\rule{0.44em}{0ex}}---------\left(iv\right)\phantom{\rule{0.22em}{0ex}}$ (i) and (iv) are equal Q.12.15 The total energy of an electron in the first excited state of the hydrogen atom is about - 3.4 eV.(a) What is the kinetic energy of the electron in this state?(b) What is the potential energy of the electron in this state?(c) Which of the answers above would change if the choice of the zero of potential energy is changed? A.12.15(a)Kinetic energy of the electron (Hydrogen Atom)K=12mv2=e28𝜋𝜖0r$K=\frac{1}{2}m{v}^{2}=\frac{{e}^{2}}{8𝜋{𝜖}_{0}r}$rn=(n2m)(h2𝜋)24𝜋𝜖0e2 ${r}_{n}=\left(\frac{{n}^{2}}{m}\right)\left(\frac{h}{2𝜋}{\right)}^{2}\frac{4𝜋{𝜖}_{0}}{{e}^{2}}\phantom{\rule{0.44em}{0ex}}$ So , K=me48 n2 𝜖20 h2 =13.6n2 eV $So\phantom{\rule{0.22em}{0ex}},\phantom{\rule{0.22em}{0ex}}K=\frac{m{e}^{4}}{8\phantom{\rule{0.22em}{0ex}}{n}^{2}\phantom{\rule{0.22em}{0ex}}{𝜖}_{0}^{2}\phantom{\rule{0.22em}{0ex}}{h}^{2}}\phantom{\rule{0.22em}{0ex}}=\frac{13.6}{{n}^{2}}\phantom{\rule{0.22em}{0ex}}eV\phantom{\rule{0.22em}{0ex}}$ first excited state means , n=2K=13.6n2 eV=13.64 eV=3.4 eV $K=\frac{13.6}{{n}^{2}}\phantom{\rule{0.22em}{0ex}}eV=\frac{13.6}{4}\phantom{\rule{0.22em}{0ex}}eV=3.4\phantom{\rule{0.22em}{0ex}}eV\phantom{\rule{0.44em}{0ex}}$ (b)Potential energy=U=- e24𝜋𝜖0r=-2K$Potential\phantom{\rule{0.22em}{0ex}}energy=U=-\phantom{\rule{0.22em}{0ex}}\frac{{e}^{2}}{4𝜋{𝜖}_{0}r}=-2K$ Potential energy=U=-2 ×3.4=-6.8 eV$Potential\phantom{\rule{0.22em}{0ex}}energy=U=-2\phantom{\rule{0.22em}{0ex}}×3.4=-6.8\phantom{\rule{0.22em}{0ex}}eV$ (d) Kinetic energy will not change if choice of the zero of potential energy is changed.Potential energy will change if choice of the zero of potential energy is changed. Q.12.16 If Bohr’s quantisation postulate (angular momentum = nh2𝜋$\frac{nh}{2𝜋}$) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun? A.12.16 Angular momentum of planet earth is about = 2.663×1040 kg m2 s-1$2.663×{10}^{40}\phantom{\rule{0.22em}{0ex}}kg\phantom{\rule{0.22em}{0ex}}{m}^{2}\phantom{\rule{0.22em}{0ex}}{s}^{-1}$ So,2.663×1040=$2.663×{10}^{40}=$nh2𝜋$\frac{nh}{2𝜋}$ n=$n=$2.663×1040 ×2𝜋6.63×10-34$\frac{2.663×{10}^{40}\phantom{\rule{0.22em}{0ex}}×2𝜋}{6.63×{10}^{-34}}$ n=2.522×1074$n=2.522×{10}^{74}$ For large value of n, successive successive energies and angular momentum of the quantised levels are very small, so levels can be considered as continuous. Q.12.17 Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (𝜇$𝜇$) of mass about 207 me$207\phantom{\rule{0.22em}{0ex}}{m}_{e}$ orbits around a proton]. A.12.17 Electron Orbits for Hydrogen atom then207mev2er=14𝜋𝜖0e2r2 $\frac{207{m}_{e}{v}_{e}^{2}}{r}=\frac{1}{4𝜋{𝜖}_{0}}\frac{{e}^{2}}{{r}^{2}}\phantom{\rule{0.66em}{0ex}}$⟹ 207mev2e=14𝜋𝜖0e2r ---------- (i) $⟹\phantom{\rule{0.22em}{0ex}}207{m}_{e}{v}_{e}^{2}=\frac{1}{4𝜋{𝜖}_{0}}\frac{{e}^{2}}{r}\phantom{\rule{0.66em}{0ex}}----------\phantom{\rule{0.22em}{0ex}}\left(i\right)\phantom{\rule{0.22em}{0ex}}$ 207 me ve r=h2𝜋$\phantom{\rule{0.22em}{0ex}}207\phantom{\rule{0.22em}{0ex}}{m}_{e\phantom{\rule{0.22em}{0ex}}}{v}_{e}\phantom{\rule{0.22em}{0ex}}r=\frac{h}{2𝜋}$ ⟹ve =h2𝜋 207 me r ---------- (ii) $⟹{v}_{e}\phantom{\rule{0.22em}{0ex}}=\frac{h}{2𝜋\phantom{\rule{0.22em}{0ex}}207\phantom{\rule{0.22em}{0ex}}{m}_{e}\phantom{\rule{0.22em}{0ex}}r\phantom{\rule{0.22em}{0ex}}}\phantom{\rule{2.86em}{0ex}}----------\phantom{\rule{0.22em}{0ex}}\left(ii\right)\phantom{\rule{0.22em}{0ex}}$ By (i) and (ii) 207 me h24𝜋2 (207)2 m2e r2 =14𝜋𝜖0e2r$\frac{\phantom{\rule{0.22em}{0ex}}207\phantom{\rule{0.22em}{0ex}}{m}_{e}\phantom{\rule{0.22em}{0ex}}{h}^{2}}{4{𝜋}^{2}\phantom{\rule{0.22em}{0ex}}\left(207{\right)}^{2}\phantom{\rule{0.22em}{0ex}}{m}_{e}^{2}\phantom{\rule{0.22em}{0ex}}{r}^{2}\phantom{\rule{0.22em}{0ex}}}=\frac{1}{4𝜋{𝜖}_{0}}\frac{{e}^{2}}{r}$ ⟹ r=h2 4𝜋𝜖0 4𝜋2 207 me e2$⟹\phantom{\rule{0.22em}{0ex}}r=\frac{{h}^{2}\phantom{\rule{0.22em}{0ex}}4𝜋{𝜖}_{0}\phantom{\rule{0.44em}{0ex}}}{4{𝜋}^{2}\phantom{\rule{0.22em}{0ex}}207\phantom{\rule{0.22em}{0ex}}{m}_{e}^{}\phantom{\rule{0.22em}{0ex}}{e}^{2}}$ ⟹ r=(6.63×10-34)2 4𝜋2 (9.1×10-31) (207) (1.6×10-19)2×(9×109) $⟹\phantom{\rule{0.22em}{0ex}}r=\frac{\left(6.63×{10}^{-34}{\right)}^{2}\phantom{\rule{0.66em}{0ex}}}{4{𝜋}^{2}\phantom{\rule{0.22em}{0ex}}\left(9.1×{10}^{-31}\right)\phantom{\rule{0.22em}{0ex}}\left(207\right)\phantom{\rule{0.22em}{0ex}}\left(1.6×{10}^{-19}{\right)}^{2}×\left(9×{10}^{9}\right)}\phantom{\rule{0.22em}{0ex}}$ ⟹ r=0.000025681×10-18 m =2.5681×10-13 m $⟹\phantom{\rule{0.22em}{0ex}}r=0.000025681×{10}^{-18}\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}=2.5681×{10}^{-13}\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}$ ⟹ r=2.5681×10-13 m $⟹\phantom{\rule{0.22em}{0ex}}r=2.5681×{10}^{-13}\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}$