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=14πœ‹πœ–0(2e)(Ze)r2
Electron Orbits for Hydrogen atommv2r=14πœ‹πœ–0e2r2Orbirradius=r=e24πœ‹πœ–0mv2
Kinetic energy of the electron (Hydrogen Atom)K=12mv2=e28πœ‹πœ–0r
Potential energy of the electron (Hydrogen Atom)U=-e24πœ‹πœ–0r
Total energy of the electron (Hydrogen Atom)E=K+U=-e28πœ‹πœ–0r
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 h2πœ‹ where h is the Planck’s constant (=6.6Γ—10-34Js). (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 orbit (for hydrogen):rn=(n2m)(h2πœ‹)24πœ‹πœ–0e2
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 =∞ and has an energy of 0 eV.
Total Energy of the electron in nth orbit for hydrogen atomEn=-me48n2πœ–02h2 En=-2.18Γ—10-18n2J En=-13.6n2eV
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 to the lower energy state nf the difference of energy is carried away by photon of frequency 𝜈ifh𝜈if=Eni-Enf⟹h𝜈if=me48πœ–02h2(1nf2-1ni2)⟹𝜈if=me48πœ–02h3(1nf2-1ni2)RydbergConstant=R=me48πœ–02h3c
Lyman Series : 𝜈=Rc(112-1n2) where , n = 2, 3, 4, 5 . .
Balmer Series : 𝜈=Rc(122-1n2) where , n = 3, 4, 5 . .
Paschen Series : 𝜈=Rc(132-1n2) where , n = 4, 5 . .
Brackett Series : 𝜈=Rc(142-1n2) where , n = 5, 6 . .
Pfund Series : 𝜈=Rc(152-1n2) where , n = 6, 7, 8 . .
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. 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) where , n = 4, 5 . . For shortest wavelength 𝜈(frequency) must be maximum. this will happen when n=infinity𝜈=1.097Γ—107Γ—cΓ—(132-1∞2)=0.122Γ—107c⟹𝜈=0.122Γ—107cβŸΉπœ†=c𝜈=10.122Γ—107=8.196Γ—10-7=8.196Γ—10-9βŸΉπœ†=819.6nm
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 byπ›₯E=h𝜈π›₯E=2.3eV=2.3Γ—1.6Γ—10-19Joulesh=6.626Γ—10-34m2kgs-1𝜈=π›₯Eh=2.3Γ—1.6Γ—10-166.626Γ—10-34=5.554Γ—1014Hz
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=-e24πœ‹πœ–0r Kinetic energy K=e28πœ‹πœ–0r So, let Kinetic enegy = x then Poential energy = -2x ⟹x+(-2x)=-13.6⟹-x=-13.6⟹x=13.6⟹-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 1πœ†=R(112-142)=1.097Γ—107(11-116)1πœ†=1.097Γ—107(1516)πœ†=0.9723Γ—10-7=97.23nmfrequency=𝜈=cπœ†=3Γ—1080.9723Γ—10-7frequency=𝜈=3.1Γ—1015Hz
Q.12.8The radius of the innermost electron orbit of a hydrogen atom is 5.3Γ—10-11m. What are the radii of the n = 2 and n =3 orbits?A.12.8As per Bohr's principal, Radius of nth orbit:rn=(n2m)(h2πœ‹)24πœ‹πœ–0e2 So,rn∝n2 r1r2=n12n22⟹5.3Γ—10-11r2=1222⟹r2=21.2Γ—10-11=2.12Γ—10-10m r1r3=n12n32⟹5.3Γ—10-11r3=1232⟹r3=47.7Γ—10-11=4.77Γ—10-10m
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=-2.18Γ—10-18n2Joules=-13.6n2eV E1=-13.6eVEnergyofelectronbeam=12.5eVSo,energyoflydrogenbecomes=-13.6+12.5=-1.1eV En=-13.6n2eV E1=-13.6eVE2=-3.4eVE3=-1.5eVE4=-1.01eV As, E4>-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 series1πœ†=R(122-132)=1.097Γ—107(14-19)⟹1πœ†=1.522Γ—106mβŸΉπœ†=656.3nm From 3 to 1 will be Lyman series1πœ†=R(112-132)=1.097Γ—107(11-19)⟹1πœ†=9.751Γ—106mβŸΉπœ†=102.55nmβ‰ˆ103nm From 2 to 1 will be Lyman series1πœ†=R(112-122)=1.097Γ—107(11-14)⟹1πœ†=8.2275Γ—106mβŸΉπœ†=121.55nmβ‰ˆ122nm
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=me48πœ–02h3(1nf2-1ni2) when hydrogen atom de-excites from level n to level (n–1) then 𝜈=me48πœ–02h3(1(n-1)2-1n2)=me48πœ–02h3(2n-1n2(n-1)2) considering n is very large πœˆβ‰ˆme48πœ–02h3(2nn4)πœˆβ‰ˆme48πœ–02h3(2n3)--------------(i) Now meve2r=14πœ‹πœ–0e2r2⟹meve2=14πœ‹πœ–0e2r----------(ii) mever=nh2πœ‹ ⟹ve=nh2πœ‹mer----------(iii) By (i) and (ii) men2h24πœ‹2me2r2=14πœ‹πœ–0e2r ⟹r=n2h2πœ–0πœ‹mee2 Frequency of revolution of electron=𝜈e 𝜈e=πœ”2πœ‹=ve2πœ‹r=nh4πœ‹2mer2=nh4πœ‹2meπœ‹2me2e4n4h4πœ–02 𝜈e=mee44h3n3πœ–02=me48πœ–02h3(2n3)---------(iv) (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=e28πœ‹πœ–0rrn=(n2m)(h2πœ‹)24πœ‹πœ–0e2 So,K=me48n2πœ–02h2=13.6n2eV first excited state means , n=2K=13.6n2eV=13.64eV=3.4eV (b)Potentialenergy=U=-e24πœ‹πœ–0r=-2K Potentialenergy=U=-2Γ—3.4=-6.8eV (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œ‹) 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 = 2.663Γ—1040kgm2s-1 So,2.663Γ—1040=nh2πœ‹ n=2.663Γ—1040Γ—2πœ‹6.63Γ—10-34 n=2.522Γ—1074 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 (πœ‡) of mass about 207me orbits around a proton]. A.12.17 Electron Orbits for Hydrogen atom then207meve2r=14πœ‹πœ–0e2r2⟹207meve2=14πœ‹πœ–0e2r----------(i) 207mever=h2πœ‹ ⟹ve=h2πœ‹207mer----------(ii) By (i) and (ii) 207meh24πœ‹2(207)2me2r2=14πœ‹πœ–0e2r ⟹r=h24πœ‹πœ–04πœ‹2207mee2 ⟹r=(6.63Γ—10-34)24πœ‹2(9.1Γ—10-31)(207)(1.6Γ—10-19)2Γ—(9Γ—109) ⟹r=0.000025681Γ—10-18m=2.5681Γ—10-13m ⟹r=2.5681Γ—10-13m