Chapter 27 Atomic Structure PDF

60 Atomic structure 49 Important Atomic Models. E3 (1) Thomson's model J.J. Thomson gave the first idea regarding structure of atom. According to this model. Positively charged sphere – – ID (i) An atom is a solid sphere in which entire and positive charge and it's mass is uniformly distributed and...

60 Atomic structure 49 Important Atomic Models. E3 (1) Thomson's model J.J. Thomson gave the first idea regarding structure of atom. According to this model. Positively charged sphere – – ID (i) An atom is a solid sphere in which entire and positive charge and it's mass is uniformly distributed and in which negative charge (i.e. electron) are embedded like seeds in watermelon. – – – Electron D YG U – Success and failure Explained successfully the phenomenon of thermionic emission, photoelectric emission and ionization. The model fail to explain the scattering of - particles and it cannot explain the origin of spectral lines observed in the spectrum of hydrogen and other atoms. (2) Rutherford's model U Rutherford's -particle scattering experiment ST Rutherford performed experiments on the scattering of alpha particles by extremely thin gold foils and made the following observations  Number of scattered particles : b r0 N N 1 sin 4 ( / 2) Nucleus N(180° ) -particle (energy E ) (i) Most of the -particles pass through the foil straight away undeflected. (ii) Some of them are deflected through small angles.  50 Atomic Structure (iii) A few -particles (1 in 1000) are deflected through the angle more than 90 o. (iv) A few  -particles (very few) returned back i.e. deflected by 180o. (v) Distance of closest approach (Nuclear dimension) 60 The minimum distance from the nucleus up to which the -particle approach, is called the 1 Ze 2 1. ; E  mv 2  K.E. of -particle distance of closest approach (r0). From figure r0  4 0 E 2 b is the thickness of the foil and N is the number of -particles scattered in a ID Note :  If t Ze 2 cot( / 2)  b  cot( / 2) 1 2 4 0  mv  2  E3 (vi) Impact parameter (b) : The perpendicular distance of the velocity vector ( v ) of the particle from the centre of the nucleus when it is far away from the nucleus is known as impact parameter. It is given as particular direction ( = constant), it was observed that N t N  constant  1  1. N 2 t2 t U After Rutherford's scattering of -particles experiment, following conclusions were made as regard as atomic structure : –15 10 D YG (a) Most of the mass and all of the charge of an atom concentrated in a very small region is called atomic nucleus. (b) Nucleus is positively charged and it's size is of the order of m  1 Fermi. + Nucleus 10–15 m 10–10 m Size of the nucleus = 1 Fermi = 10 – 15 m Size of the atom 1 Å = 10–10 m U (c) In an atom there is maximum empty space and the electrons revolve around the nucleus in the same way as the planets revolve around the sun. Atom Draw backs ST (i) Stability of atom : It could not explain stability of atom because according to classical electrodynamic theory an accelerated charged particle should continuously radiate energy. Thus an electron moving in an circular path around the nucleus should also radiate energy and thus move into smaller and smaller orbits of e– gradually decreasing radius and it should ultimately fall into nucleus. (ii) According to this model the spectrum of atom must be continuous where as practically it is a line spectrum. (iii) It did not explain the distribution of electrons outside the nucleus. (3) Bohr's model Instability of atom Atomic structure 51 Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom) Bohr's model is based on the following postulates. 60 (i) The electron can revolve only in certain discrete non-radiating orbits, called stationary orbits, for which total angular momentum of the revolving electrons is an integral multiple of h ( ) 2 E3  h  i.e. L  n    mvr ; where n = 1, 2, 3, ……..= Principal quantum number  2  (ii) The radiation of energy occurs only when an electron jumps from one permitted orbit to another. D YG Emission U E1 – E2 = h E2 ID When electron jumps from higher energy orbit (E1) to lower energy orbit (E2) then difference of energies of these orbits i.e. E1 – E2 emits in the form of photon. But if electron goes from E2 to E1 it absorbs E1 E1 the same amount of energy. Note :  According to Bohr theory the momentum of an atom will be E1 – E2 = h E2 Absorptio n e  revolving in second orbit of H 2 h   For an electron in the nth orbit of hydrogen atom in Bohr model, circumference of orbit  n  ; where  = de-Broglie wavelength. Bohr's Orbits (For Hydrogen and H2-Like Atoms). U (1) Radius of orbit ST For an electron around a stationary nucleus the electrostatics force of attraction provides the necessary centripetal force i.e. (Ze )e mv 2  4 0 r 2 r 1 also mvr  ……. (i) nh 2 …….(ii) From equation (i) and (ii) radius of nth orbit n 2 h 2 0 n 2h2 n2   0. 53 rn  Å Z 4 2 kZme 2 mZe 2  rn  n2 Z  1  where k   4 0   r n 52 Atomic Structure Note :  The radius of the innermost orbit (n = 1) hydrogen atom (z = 1) is called Bohr's radius a0 i.e. a 0  0.53 Å. (2) Speed of electron 2kZe 2 Ze 2  c   nh 2 0 nh  137 v  Z 6 Z m / sec .  2.2  10 n  n where (c = speed of light 3  108 m/s) E3 vn  60 From the above relations, speed of electron in nth orbit can be calculated as n Note :  The ratio of speed of an electron in ground state in Bohr's first orbit of hydrogen e2 1  2 0 ch 137 (where c = speed of light in ID atom to velocity of light in air is equal to air) (3) Some other quantities Quantity Formula n  (2) Frequency (3) Time period U (4) Angular momentum ST (5) Corresponding current (6) Magnetic moment (7) Magnetic field (4) Energy n  Tn  Dependency on n and Z vn mz e  rn 2 02 n 3 h 3 n  n mz 2 e 4  2 4  02 n 3 h 3 n  2 4 D YG (1) Angular speed U For the revolution of electron in nth orbit, some other quantities are given in the following table 1 n  4  02 n 3 h 3 Tn  mz 2 e 4 mz 2 e 5 in  4  02 n 3 h 3   Z2 n3 n3 Z2 Z2 n3 Mn  n M n  in A  in  rn2 0  (where n3 Ln  n  h  L n  mv n rn  n    2  in  e  n  Z2 eh  Bohr 4 m magneton) B  0 in 2rn  m 2 z 3 e 7  0 8  03 n 5 h 5 B Z3 n5 Atomic structure 53 (i) Potential energy : An electron possesses some potential energy because it is found in the field of nucleus potential energy of electron in nth orbit of radius rn is given by (Ze )(e ) kZe 2 U  k.  rn rn As we know k Ze 2 | U | mv 2 k. (Ze )(e ) K    Kinetic energy  2rn 2 rn rn2 60 (ii) Kinetic energy : Electron posses kinetic energy because of it's motion. Closer orbits have greater kinetic energy than outer ones.  me 4 E    2 2  8 0 h k Ze 2 2rn  me 4  z2  .    8  2 ch 3  n2  0  where R  also rn  n 2 h 2 0 mze 2. Hence  z2 Z2 Z2  ch   R ch   13. 6 eV  n2 n2 n2  ID E  E3 (iii) Total energy : Total energy (E) is the sum of potential energy and kinetic energy i.e. E =K+U me 4 = Rydberg's constant = 1.09  107 per metre 8  02 ch 3 U Note :  Each Bohr orbit has a definite energy 13.6 eV n2  The state with n = 1 has the lowest (most negative) energy. For hydrogen atom it is E1 = – 13.6 eV.  Rch = Rydberg's energy ~– 2.17  10 18 J ~– 31.6 eV. D YG  For hydrogen atom (Z = 1)  En   U. 2 (iv) Ionisation energy and potential : The energy required to ionise an atom is called ionisation energy. It is the energy required to make the electron jump from the present orbit to the infinite orbit. U  E  K  ST  Z2  13.6 Z 2 Hence E ionisation  E   E n  0    13.6 2    eV n  n2   13.6(1)2  13.6 eV n2 The potential through which an electron need to be accelerated so that it acquires energy E equal to the ionisation energy is called ionisation potential. Vionisation  ionisation e (v) Excitation energy and potential : When the electron is given energy from external source, it jumps to higher energy level. This phenomenon is called excitation. For H2-atom in the ground state Eionisation  The minimum energy required to excite an atom is called excitation energy of the particular excited state and corresponding potential is called exciting potential. 54 Atomic Structure E Excitation  EFinal  EInitial and VExcitation  E excitation e n 13.6. (B.E.) E3 Note :  For hydrogen atom principle quantum number 60 (vi) Binding energy (B.E.) : Binding energy of a system is defined as the energy released when it's constituents are brought from infinity to form the system. It may also be defined as the energy needed to separate it's constituents to large distances. If an electron and a proton are initially at rest and brought from large distances to form a hydrogen atom, 13.6 eV energy will be released. The binding energy of a hydrogen atom is therefore 13.6 eV. (5) Energy level diagram The diagrammatic description of the energy of the electron in different orbits around the nucleus is called energy level diagram. Infinit e Fourth n=4 n=3 Third n=2 Second First E = 0 eV Third E4 = – 0.85 eV Second E3 = – 1.51 eV First Ground E2 = – 3.4 eV E1 = – 13.6 eV D YG n=1 Infinite U n= ID Energy level diagram of hydrogen/hydrogen like atom Principl e quantu m number Excited state Orbit 0 eV 0 eV – 0.85 Z2 – 1.51 Z – 3.4 Z + 0.85 eV 2 + 1.51 eV 2 + 3.4 eV – 13.6 Z 2 + 13.6 eV Energy for H2 Energy for H2 – atom – like atom Note :  In hydrogen atom excitation energy to excite electron Ionisation energy from this level (for H2 – atom) from ground state to first U excited state will be  3.4  (13.6)  10.2 eV. and from ground state to second excited state it is [  1.51  (13.6)  12.09 eV ]. ST  In an H 2 atom when e  makes a transition from an excited state to the ground state it’s kinetic energy increases while potential and total energy decreases. (6) Transition of electron When an electron makes transition from higher energy level having energy E2(n2) to a lower energy level having energy E1 (n1) then a photon of frequency  is emitted (i) Energy of emitted radiation E  E 2  E1   Rc h Z 2 n 22  Rch Z 2    n12  (ii) Frequency of emitted radiation      13. 6 Z 2  1  1    n2 n2  2    1 E2 n2 E1 n1 E,, Emission Atomic structure 55 E  h     1 E E 2  E1 1    Rc Z 2  2  2  h h  n1 n 2  (iii) Wave number/wavelength  number is the  1 1  13.6 Z  RZ 2  2  2   hc   n1 n 2  1 2 number of waves in unit length   1  60 Wave  1 1    n2  n2  2   1   c E3 (iv) Number of spectral lines : If an electron jumps from higher energy orbit to lower energy orbit it emits raidations with various spectral lines. NE  (n 2  n1  1)(n 2  n1 ) 2 ID If electron falls from orbit n2 to n1 then the number of spectral lines emitted is given by U If electron falls from nth orbit to ground state (i.e. n2 = n and n1 = 1) then number of spectral lines n (n  1) emitted N E  2 Note :  Absorption spectrum is obtained only for the transition from lowest energy level to D YG higher energy levels. Hence the number of absorption spectral lines will be (n – 1). (v) Recoiling of an atom : Due to the transition of electron, photon is emitted and the atom is recoiled Recoil momentum of atom = momentum of photon  U Also recoil energy of atom  p2 h2  2m 2m  2  1 1   hRZ 2  2  2    n1 n 2  h (where m = mass of recoil atom) (7) Draw backs of Bohr's atomic model ST (i) It is valid only for one electron atoms, e.g. : H, He+, Li+2, Na+1 etc. (ii) Orbits were taken as circular but according to Sommerfield these are elliptical. (iii) Intensity of spectral lines could not be explained. (iv) Nucleus was taken as stationary but it also rotates on its own axis. (v) It could not be explained the minute structure in spectrum line. (vi) This does not explain the Zeeman effect (splitting up of spectral lines in magnetic field) and Stark effect (splitting up in electric field) (vii) This does not explain the doublets in the spectrum of some of the atoms like sodium (5890Å & 5896Å) 56 Atomic Structure Hydrogen Spectrum and Spectral Series. When hydrogen atom is excited, it returns to its normal unexcited (or ground state) state by emitting the energy it had absorbed earlier. This energy is given out by the atom in the form of radiations of different wavelengths as the electron jumps down from a higher to a lower 60 orbit. Transition from different orbits cause different wavelengths, these constitute spectral series which are characteristic of the atom emitting them. When observed through a spectroscope, these radiations are imaged as sharp and straight vertical lines of a single colour. + + + ID Emission spectra Spectrum E3 Photon of wavelength  Spectral series The spectral lines arising from the transition of electron forms a spectra series. U (i) Mainly there are five series and each series is named after it's discover as Lymen series, Balmer series, Paschen series, Bracket series and Pfund series. D YG (ii) According to the Bohr's theory the wavelength of the radiations emitted from hydrogen atom is given by 1 1   R 2  2   n1 n 2  1 ST U where n2 = outer orbit (electron jumps from this orbit), n1 = inner orbit (electron falls in this orbit) (iii) First line of the series is called first member, for this line wavelength is maximum (max) Atomic structure 57 (iv) Last line of the series (n2 = ) is called series limit, for this line wavelength is minimum (min) Transition (n 22  n12 )R Maximum wavelength n1  n and n 2  n  1  m ax  n 2 (n  1) 2 (2n  1)R  max  (1) 2 (1  1) 2 4  (2  1  1)R 3R  min  n2 R  max (n  1) 2   min (2n  1) n1 = n = 1 1 m in  R 2.Balmer series n2 = 3, 4, 5 … n1 = 2 n1 = n = 2, n2 = 2 + 1 = 3 36 m ax  5R  m in  3. Paschen series n2 = 4, 5, 6 … n1 = 3 n1 = n = 3, n2 = 3 + 1 = 4 144 m ax  7R 4. Bracket series n2 = 5, 6, 7 …  n1 = 4 n1 = n = 4, n2 = 4 + 1 = 5 400 m ax  9R n2 = 6, 7, 8 …  n1 = 5 n1 =  = 5, n2 = 5 + 1 = 6 900 m ax  11 R Ultraviolet region 9 5 Visible region n1 = n = 3 9 m in  R 16 7 Infrared region n1 = n = 4 16 m in  R 25 9 Infrared region 36 11 Infrared region 4 R U D YG Region 4 3 ID n2 = 2, 3, 4 … n1 = 1 Pfund 2    1  n1  R  n 22   Minimum wavelength n 2  , n1  n 1. Lymen series 5. series n12 E3 Spectral series n12 n 22 60 Wavelength ()  m in  25 R Quantum Numbers. U An atom contains large number of shells and subshells. These are distinguished from one another on the basis of their size, shape and orientation (direction) in space. The parameters are expressed in terms of different numbers called quantum number. ST Quantum numbers may be defined as a set of four number with the help of which we can get complete information about all the electrons in an atom. It tells us the address of the electron i.e. location, energy, the type of orbital occupied and orientation of that orbital. (1) Principal Quantum number (n) : This quantum number determines the main energy level or shell in which the electron is present. The average distance of the electron from the 1 E n  2 and rn  n 2 nucleus and the energy of the electron depends on it. (in Hn atom) The principal quantum number takes whole number values, n = 1, 2, 3, 4,…..  (2) Orbital quantum number (l) or azimuthal quantum number (l) 58 Atomic Structure This represents the number of subshells present in the main shell. These subsidiary orbits within a shell will be denoted as 1, 2, 3, 4 … or s, p, d, f … This tells the shape of the subshells. The orbital angular momentum of the electron is given as L  l(l  1) h 2 (for a particular value of n). 60 For a given value of n the possible values of l are l = 0, 1, 2, ….. upto (n – 1) E3 (3) Magnetic quantum number (ml) : An electron due to it's angular motion around the nucleus generates an electric field. This electric field is expected to produce a magnetic field. Under the influence of external magnetic field, the electrons of a subshell can orient themselves in certain preferred regions of space around the nucleus called orbitals. The magnetic quantum number determines the number of preferred orientations of the electron present in a subshell. ID The angular momentum quantum number m can assume all integral value between – l to +l including zero. Thus ml can be – 1, 0, + 1 for l = 1. Total values of ml associated with a particular value of l is given by (2l + 1). U (4) Spin (magnetic) quantum number (ms) : An electron in atom not only revolves around D YG the nucleus but also spins about its own axis. Since an electron can spin either in clockwise direction or in anticlockwise direction. Therefore for any particular value of magnetic quantum 1 number, spin quantum number can have two values, i.e. m s  (Spin up) or 2 1 ms   (Spin down) 2 This quantum number helps to explain the magnetic properties of the substance. Electronic Configurations of Atoms. U The distribution of electrons in different orbitals of an atom is called the electronic configuration of the atom. The filling of electrons in orbitals is governed by the following rules. ST (1) Pauli's exclusion principle "It states that no two electrons in an atom can have all the four quantum number (n, l, ml and ms) the same." It means each quantum state of an electron must have a different set of quantum numbers n, l, ml and ms. This principle sets an upper limit on the number of electrons that can occupy a shell. N max in one shell = 2n2; Thus Nmax in K, L, M, N …. shells are 2, 8, 18, 32, Note :  The maximum number of electrons in a subshell with orbital quantum number l is 2(2l + 1). (2) Aufbau principle Atomic structure 59 Electrons enter the orbitals of lowest energy first. As a general rule, a new electron enters an empty orbital for which (n + l ) is minimum. In case the value (n  l) is equal for two orbitals, the one with lower value of n is filled first. 60 Thus the electrons are filled in subshells in the following order (memorize) 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, …… (3) Hund's Rule E3 When electrons are added to a subshell where more than one orbital of the same energy is available, their spins remain parallel. They occupy different orbitals until each one of them has at least one electron. Pairing starts only when all orbitals are filled up. ID Pairing takes place only after filling 3, 5 and 7 electrons in p, d and f orbitals, respectively. Concepts With the increase in principal quantum number the energy difference between the two successive energy level decreases, while wavelength of spectral line increases. E'  E' '  E' ' ' '  ' '  ' ' ' E,  E,  D YG E  E' E' ' E' ' ' 1 1 1 1     ' ' ' ' ' ' U  E,  n=4 n=3 n=2 E,  n=1 Rydberg constant is different for different elements R( =1.09  107 m–1) is the value of Rydberg constant when the nucleus is considered to be infinitely massive as compared to the revolving electron. In other words, the nucleus is considered to be stationary. In case, the nucleus is not infinitely massive or stationary, then the value of Rydberg constant is given as R where m is the mass of electron and M is the mass of nucleus. R' m 1 M  Atomic spectrum is a line spectrum Each atom has it's own characteristic allowed orbits depending upon the electronic configuration. Therefore photons emitted during transition of electrons from one allowed orbit to inner allowed orbit are of some definite energy only. They do not have a continuous graduation of energy. Therefore the spectrum of the emitted light has only some definite lines and therefore atomic spectrum is line spectrum. ST U   Just as dots of light of only three colours combine to form almost every conceivable colour on T.V. screen, only about 100 distinct kinds of atoms combine to form all the materials in the universe. Example s Example: 1 The ratio of areas within the electron orbits for the first excited state to the ground state for hydrogen atom is (a) 16 : 1 Solution : (a) For a hydrogen atom (b) 18 : 1 (c) 4 : 1 (d) 2 : 1 60 Atomic Structure r12 Radius r  n 2  Example: 2 r22  n14 n 24  r12 r22  n14 n 24  A1 n4 A 24 16  14  4  16  1  A2 A 1 n2 1 2 The electric potential between a proton and an electron is given by V  V0 ln r , where r0 is a r0 60 constant. Assuming Bohr’s model to be applicable, write variation of rn with n, n being the principal quantum number [IIT-JEE (Screening) 2003] Potential energy U  eV  eV 0 ln  Force F   (d) rn  1 / n 2 r r0 E3 Solution : (a) (c) rn  n (b) rn  1 / n (a) rn  n 2 dU eV0 . The force will provide the necessary centripetal force. Hence dr r eV 0 mv 2  v  r r eV 0 m …..(i) mvr  and nh 2 …..(ii) The innermost orbit of the hydrogen atom has a diameter 1.06 Å. The diameter of tenth orbit is (a) 5.3 Å U Example: 3 ID  nh  m Dividing equation (ii) by (i) we have mr   or r  n   2  eV 0 (b) 10.6 Å 2 (c) 53 Å 2 (d) 106 Å d2 d 2  n2   10       1.06  1  d 1  n1  D YG r2  n 2    r1  n1  [UPSEAT 2002] 2 Solution : (d) Using r  n 2  Example: 4 Energy of the electron in nth orbit of hydrogen atom is given by En   or  d = 106 Å 13.6 eV. The amount of n2 energy needed to transfer electron from first orbit to third orbit is (a) 13.6 eV Using E   13.6 n2 For n = 1 , E1  (c) 12.09 eV (d) 1.51 eV eV 13.6 U Solution : (c) (b) 3.4 eV 1 2  13.6 eV and for n = 3 E 3   13.6 32  1.51 eV ST So required energy  E 3  E1  1.51  (13.6)  12.09 eV Example: 5 If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li   is (a) 122.4 eV Solution : (b) (b) 30.6 eV Using E n   13.6  Z 2 n2 (c) 13.6 eV (d) 3.4 eV eV For first excited state n = 2 and for Li  , Z = 3  E Li Example: 6  13.6 2  32   2 be 30.6 eV 13.6  9  30.6 eV. Hence, remove the electron from the first excited state of 4 The ratio of the wavelengths for 2 1 transition in Li++, He+ and H is (a) 1 : 2 : 3 (b) 1 : 4 : 9 (c) 4 : 9 : 36 [UPSEAT 2003] (d) 3 : 2 : 1 Atomic structure 61  1 1 1 1 1 1   RZ 2  2  2     2   Li :  He  :  H  : :  4 : 9 : 36   9 4 1  Z  n1 n 2  1 Solution : (c) Using Example: 7 Energy E of a hydrogen atom with principal quantum number n is given by E  13. 6 eV. The n2 60 energy of a photon ejected when the electron jumps n = 3 state to n = 2 state of hydrogen is approximately [CBSE PMT/PDT Screening 2004] (a) 1.9 eV 1  5  1 E  13. 6  2  2   13. 6   1. 9 eV 36 3  2 In the Bohr model of the hydrogen atom, let R, v and E represent the radius of the orbit, the speed of electron and the total energy of the electron respectively. Which of the following quantity is proportional to the quantum number n (a) R/E Solution : (d) (d) 3.4 eV E3 Example: 8 (c) 0.85 eV (b) E/v Rydberg constant R  mZe 2 (d) vR mZ 2 e 4 Ze 2 and energy E   2 2 2 2 0 nh 8 0 n h U Velocity v  0n2h2 (c) RE ID Solution : (a) (b) 1.5 eV Now, it is clear from above expressions R.v  n The energy of hydrogen atom in nth orbit is En, then the energy in nth orbit of singly ionised helium atom will be D YG Example: 9 (a) 4En (b) En/4 13. 6 Z 2   Z EH   H E He  Z He 2 (c) 2En (d) En/2  1      E He  4 E n. 2  2 Solution : (a) By using E   Example: 10 The wavelength of radiation emitted is 0 when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be 16 0 25 U (a) Solution : (b) n2 (b) 20 0 27 (c) 27 0 20 1 1 1  5 1   1 1 1  R  R 2  2   R     R 2  2   0  3   4 9  36 2  n1 n 2  …..(i) 1 1  1  3R  1 1  R 2  2   R    4   4 16  16 2 …..(ii) ST and From equation (i) and (ii) 20 20   5 R 16    0    36 3 R 27 27  If scattering particles are 56 for 90 o angle then this will be at 60 o angle (a) 224 Solution : (a) 25 0 16 Wavelength of radiation in hydrogen atom is given by 1 Example: 11 (d) (b) 256 Using Scattering formula (c) 98 (d) 108 62 Atomic Structure N  sin 4 ( / 2)  N2 N1 4  4  N 2  4 N 1  4  56  224 60 (b) 3.32  10 34 J-s (c) 1.05  10 34 J-s (d) 2.08  10 34 J-s E3 n 2 h n1 h h 6.6  10 34   L  (n 2  n1 )  (5  4 )  1.05  10  34 J-s 2 2 2 2  3.14 In hydrogen atom, if the difference in the energy of the electron in n = 2 and n = 3 orbits is E, the ionization energy of hydrogen atom is (a) 13.2 E Solution : (b)  sin 45      sin 30   Change in angular momentum L  L 2  L1  Example: 13 4 When an electron in hydrogen atom is excited, from its 4th to 5th stationary orbit, the change in angular momentum of electron is (Planck’s constant: h  6.6  10 34 J s ) (a) 4.16  10 34 J- s Solution : (c) 4   90         sin N  2   2     60    N1      sin   2    (b) 7.2 E (c) 5.6 E ID Example: 12 1   1  sin  2    2  sin   2 Energy difference between n = 2 and n = 3; (d) 3.2 E 1  5  1 1 1 E  K 2  2   K    K 3   4 9  36 2 U 1   1 Ionization energy of hydrogen atom n1  1 and n 2   ; E   K  2  2   K   1 Example: 14 …..(ii) 36 E  7.2 E 5 D YG From equation (i) and (ii) E   …..(i) In Bohr model of hydrogen atom, the ratio of periods of revolution of an electron in n = 2 and n = 1 orbits is (a) 2 : 1 (b) 4 : 1 [EAMCET (Engg.) 2000] (c) 8 : 1 (d) 16 : 1 T2 n3 23 8  23  3   1 T1 1 n1 T2  8T1. According to Bohr model time period of electron T  n 3  Example: 15 A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of required radiation for emitting electron from first to third Bohr orbit in Li   will be (Ionisation energy of hydrogen atom is 13.6 eV) U Solution : (c) ST (a) 182.51 Å Solution : (d) (b) 177.17 Å (c) 142.25 Å (d) 113.74 Å Energy of a electron in nth orbit of a hydrogen like atom is given by E n  13.6 Z2 n2 eV , and Z = 3 for Li Required energy for said transition 1   1 8  E  E 3  E1  13.6 Z 2  2  2   13.6  3 2    108.8 eV  108.8  1.6  10 19 J 3  9  1 Now using E  Example: 16 hc    6. 6  10 34  3  10 8 hc  0.11374  10 7 m    113.74 Å   E 108. 8  1.6  10 19 The absorption transition between two energy states of hydrogen atom are 3. The emission transitions between these states will be Atomic structure 63 (a) 3 (d) 6 Number of absorption lines = (n – 1)  3 = (n – 1)  n = 4 Hence number of emitted lines  Example: 17 (c) 5 n(n  1) 4 (4  1)  6 2 2 The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E/3 and 2E respectively. A photon of wavelength  is emitted for a transition 3 1. What will be the wavelength of emissions for transition 2 1 60 Solution : (d) (b) 4 [CPMT 1996] (b) 4/3 For transition 3 1 E  2 E  E  For transition 2 1 (c) 3/4 hc  4E 3 hc hc E  E    3 …..(i)  …..(ii) Hydrogen atom emits blue light when it changes from n = 4 energy level to n = 2 level. Which colour of light would the atom emit when it changes from n = 5 level to n = 2 level (a) Red (b) Yellow (c) Green (d) Violet In the transition from orbits 5 2 more energy will be liberated as compared to transition U Solution : (d) hc ID From equation (i) and (ii)    3  Example: 18  E  (d) 3 E3 Solution : (d) (a) /3 from 4 2. So emitted photon would be of violet light. A single electron orbits a stationary nucleus of charge +Ze, where Z is a constant. It requires 47.2 eV to excited electron from second Bohr orbit to third Bohr orbit. Find the value of Z [IIT-JEE 1981] D YG Example: 19 (a) 2 Solution : (b) (b) 5 (c) 3 (d) 4 Excitation energy of hydrogen like atom for n 2 n1  1 1  5 2 47.2  36 1   1 Z  Z2   24.98 ~ 25 E  13.6 Z 2  2  2  eV  47. 2  13. 6 Z 2  2  2   13.6  n  36 13. 6  5 2 3   n 2   1 U Z=5 The first member of the Paschen series in hydrogen spectrum is of wavelength 18,800 Å. The short wavelength limit of Paschen series is [EAMCET (Med.) 2000] ST Example: 20 (a) 1215 Å Solution : (c) (b) 6560 Å First member of Paschen series mean it's  max  (c) 8225 Å 144 7R Short wavelength of Paschen series means  min  Hence Example: 21 (d) 12850 Å 9 R  max 16 7 7    max   18 ,800  8225 Å.   min   min 7 16 16 Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is [EAMCET (Engg.) 1995; MP PMT 1997] (a) 1 : 3 (b) 27 : 5 (c) 5 : 27 (d) 4 : 9 1 For Lyman series  L1 For Balmer series  B1 1 1  3R 1  R 2  2  4 2  1 …..(i) 1  5R  1  R 2  2  3  36 2 …..(ii) From equation (i) and (ii) 5. 27 The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be (a) 3.4 eV Solution : (c)  B1  (c) 54.4 eV (d) 122.4 eV  1 1 1  1   1  1.1  10 7  Z 2  2  2   RZ 2  2  2   9    108.5  10 5  2  n1 n 2  1 Using  (b) 13.6 eV 1 108.5  10 9  1.1  10 7  Z 2  100 21  Z2  4 Z=2  9 100 108.5  10  1.1  10 7  21 ID Example: 22  L1 E3 Solution : (c) 60 64 Atomic Structure Now Energy in ground state E  13.6 Z 2 eV  13.6  2 2 eV  54.4 eV Solution : (a) U Hydrogen (H), deuterium (D), singly ionized helium (He  ) and doubly ionized lithium (Li   ) all have one electron around the nucleus. Consider n  2 to n  1 transition. The wavelengths of emitted radiations are 1, 2 , 3 and 4 respectively. Then approximately (c) 1  22  2 23  3 24 (d) 1  2  23  34 (a) 1  2  4 3  94 (b) 4 1  22  23  4 Using E  Z 2 (∵ n 1 and n 2 are same)  D YG Example: 23 hc  Z 2  Z 2  constant  1 Z12   2 Z 22  3 Z 32  4 Z 4  1  1   2  1 2  3  2 2  4  3 3   1   2  4  3  9  4. Example: 24 Hydrogen atom in its ground state is excited by radiation of wavelength 975 Å. How many lines will be there in the emission spectrum (b) 4 (c) 6 (d) 8 U (a) 2 Using 1 1 1  1   1  1.097  10 7  2  2   n = 4  R 2  2   10  975  10 n  1  n1 n 2  1 ST Solution : (c) Now number of spectral lines N  Example: 25 A photon of energy 12.4 eV is completely absorbed by a hydrogen atom initially in the ground state so that it is excited. The quantum number of the excited state is (a) n =1 Solution : (c) (b) n= 3 (c) n = 4 (d) n =  Let electron absorbing the photon energy reaches to the excited state n. Then using energy conservation   13.6 n Example: 26 n(n  1) 4 (4  1)  6. 2 2 2  13.6  12.4   13.6 n 2  1.2  n 2  13.6  12  n = 3.46 ≃ 4 1.2 The wave number of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is 20,397 cm–1. The wave number of the energy for the same transition in He  is [Haryana PMT 2000] Atomic structure 65 (a) 5,099 cm–1 (b) 20,497 cm–1 (c) 40,994 cm–1 (d) 81,998 cm–1 2 2  1 Z   1  Z    RZ 2  2  2     Z 2  2   2      4   2    4  81588 cm 1. n   1  Z1   1  1 n2  1 Using Example: 27 In an atom, the two electrons move round the nucleus in circular orbits of radii R and 4R. the ratio of the time taken by them to complete one revolution is (a) 1/4 Solution : (d) (b) 4/1 Time period T  n (c) 8/1 3 Z2 For a given atom (Z = constant) So T  n 3 …..(i) T1  R1  T2  R 2    3/2 …..(ii)  R     4R  3/2  1. 8 Ionisation energy for hydrogen atom in the ground state is E. What is the ionisation energy of Li   atom in the 2nd excited state (a) E (b) 3E Ionisation energy of atom in nth state E n  (c) 6E (d) 9E Z2 n2 U Solution : (a) radius R  n 2 ID Example: 28 (d) 1/8 E3  From equation (i) and (ii) T  R 3 / 2  and 60 Solution : (d) For hydrogen atom in ground state (n = 1) and Z = 1  E  E 0 D YG For Li  atom in 2nd excited state n = 3 and Z = 3, hence E   …..(i) E0 32  3 2  E0 …..(ii) From equation (i) and (ii) E   E. Example: 29 An electron jumps from n = 4 to n = 1 state in H-atom. The recoil momentum of H-atom (in eV/C) is (a) 12.75 Solution : (a) (b) 6.75 (c) 14.45 (d) 0.85 The H-atom before the transition was at rest. Therefore from conservation of momentum Photon momentum of H-atom or U Precoil momentum = Recoil E  E1 0.85 eV  (13.6 eV ) h eV   4   12.75 c c c c If elements with principal quantum number n > 4 were not allowed in nature, the number of possible elements would be ST Example: 30 [IIT-JEE 1983; CBSE PMT 1991, 93; MP PET 1999; RPET 1993, 2001; RPMT 1999, 2003; J & K CET 2004] (a) 60 Solution : (a) (b) 32 (c) 4 (d) 64 Maximum value of n = 4 So possible (maximum) no. of elements N  2  1 2  2  2 2  2  3 2  2  4 2  2  8  18  32  60. Tricky example: 1 If the atom 100 Fm 257 follows the Bohr model and the radius of 100 Fm 257 is n times the Bohr 66 Atomic Structure radius, then find n [IIT-JEE (Screening) 2003] (b) 200 m = 5 for  2   (0.53 Å)  (n  0.53 Å)  m  n  Z  100 n Fm 257 (the outermost shell) and z = 100 (5) 2 1  100 4 (d) 1/4 E3 m2 Solution : (d) (rm )    Z  (c) 4 60 (a) 100 (a) 79.0 (b) 51.8 ID An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy (in eV) required to remove both the electrons from a neutral helium atom is (c) 49.2 (d) 38.2 Solution : (a) After the removal of first electron remaining atom will be hydrogen like atom. U So energy required to remove second electron from the atom E  13.6  ST U D YG  Total energy required = 24.6 + 54.4 = 79 eV 22  54.4 eV 1