Atomic Structure Solved Problems

ATOMIC STRUCTURE SOLVED PROBLEMS Ex.2 An oil drop has 6.39 × 10– 19 C charge. Find out the number of electrons in this drop Sol. Charge on oil drop = 6.39 × 10– 19 C Now we know that 1.602 × 10– 19 C is the charge on one 1 electron  6.39 × 10– 19 C will be charge on 6.39  10 19 = = 4 electrons 1602 .  10 19 Ex.3 Find out the number of wave made by a Bohr electron in one complete revolution in its 3rdorbit of hydrogen atom Sol. We know that rn = r0 × n2  r3 = 0.529 × 10– 8 cm × (3)2 ( r0 = 0.529 × 10– 8 cm) Also we know that un = u0 n 219 .  10 8  u3  3 ( u0 = 2.19 × 108 cm sec–1) 2r3 2r3 No. of waves in one round =   h / mu 3 2r3  u 3  m = h Substituting the values of the different constants No. of waves in one round = 2  3.14  0.529  10 8 8  9  2.19  10  9.108  10 3  6.62  10  27 28 =3 Ex.4 The ionization energy of He+ is 19.6×10–18 J atom–1. The energy of the first stationary state of Li+2 will be (A) 21.2 × 10–18 J/atom (B) 44.10 × 10–18 J/atom (C) 63.2 × 10–18 J/atom (D) 84.2 × 10–18 J/atom (Ans. B) +2 Sol. E1 for Li = E1 for H × Z2 Li = E1 for H × 9 E1 for He+ = E1 for H × Z2 He = E1 for H × 4 9 or E1 for Li+2 = E for He+ 4 1 9 = 19.6 × 10–18 × 4 = 44.10 × 10–18 J/atom Ex.5 The ionization energy of hydrogen atom is 13.6 eV. What will be the ionization energy of He+ Sol. He+ is a hydrogen like species i.e. the electron is ionised from first orbit. Z 2EH  Ionization energy of He+ = n2 4  13.6 = = 54.4 eV 12 1 Ex.6 The ionization energy of H-aotm is 13.6 eV. The ionization energy of Li+2 ion will be (A) 13.6 eV (B) 27.2 eV (C) 54.4 eV (D) 122.4 eV (Ans D) Sol. E1 for Li+2 = E1 for H × Z2 [for Li, Z = 3] = 13.6 × 9 = 122.4 eV Ex.7 Which transition of the Hydrogen spectrum would have the same lenght as the Balmer transition, n = 4 to n = 2 of He+ spectrum (A) n2 = 2 to n1 = 1 (B) n2 = 3 to n1 = 1 (C) n2 = 4 to n1 = 2 (D) n2 = 5 to n1 = 3 (Ans. A) + Sol. For He ion, we have 1  LM 1  1 OP MN n n PQ L 1  1 OP 2 M N2 4 Q = RH Z 2 = RH 2 1 2 2 2 2 3 RH 4 Now for H atom 2 = …(A) LM MN OP PQ 1 1 1 = RH  2 2  n1 n2 Equating equs (A) and (B) we have 1 n12  1 n22 = …(B) 3 4 Obviously n1 = 1 and n2 = 2. Hence the transition n = 2 to n = 1 in hydrogen atom will have the same length as the transition n = 4 to n = 2 in He+ species. Ex.8 Given R = 1.0974 × 107 m–1 and h = 6.626 × 10– 34 Js. The ionization energy of one mole of Li+2 ions will be as follows (A) 11240 KJ mole–1 (B) 11180 KJ mole–1 –1 (C) 12350 KJ mole (D) 15240 KJ mole–1 (Ans B) Sol. The expression of Ionization energy is E = RZ2 hc For Li+2 ion, Z = 3, hence E = (1.0974 × 107 m–1) × (9) × (6.626 × 10–34 J.S.) × (3 × 108 ms–1) = 1.964 × 10–17 J For one mole of ions, we have E’ = NA . E = (6.023 x 1023 mol–1) (1.964 x 10–17 J) = 1.118 x 107 J mol–1 = 11180 KJ mol–1 ATOMIC STRUCTURE Ex.9 Calculate the energy emitted when electron of 1.0 g atom of hydrogen undergo transition giving the spectral line of lowest energy in the visible region of its atomic spectrum (RH = 1.1 × 107 m–1, c = 3 × 108 ms–1, h = 6.62 × 10–34 Js). Sol. The spectral line lies in the visible regionn i.e., it corresponds to the Balmer series i.e. n2 = 2 and hence n1 = 3, 4, 5, etc. For lowest energy n1 = 3 Substituting the values in the following relation. 1  RH   LM 1 MN n 2 2 1 O  P n PQ 1  2 .  10 7  = 11 LM 1  1 OP N4 9Q .  10 7  = 11 5 36 = 11 .  10 7  5 = 6.55 × 10– 7 m Now we know that c E = hv = h ×  6.62  10 34  3  10 8 6.55  10 7 1 3  10  8 OP Q  109 .  10 7  Z 2  Z2 = 4 and Z = 2  = 109678 = 911.7 × 10– 8 = 911.7 Å RH  3.03  10 19 J Ex.10 Estimate the difference in energy between 1st and 2nd Bohr orbit for a H atom. At what minimum atomic no., a transition from n = 2 to n = 1 energy level would result in the emission of X- ray with  = 3.0 × 10– 8 m. Which hydrogen spectrum like species does this atomic no. corresponds to Sol. E1 for H = – 13.6 eV  E2 for H = ( – 13.6/22) = – 13.6/4 = – 3.4 eV  E2 – E1 = – 3.4 – ( – 13.6) = + 10.2 eV Also for transition of H like atom ;  = 3.0 x 10– 8 m LM N = 2 c v=   Energy corresponding to 1g atom of hydrogen = 3.03 × 10– 19 × 6.02 × 1023 4 = 18.25 × 10 J = 182.5 KJ 1 1 1 = RH . Z2 2  2 1 2  LM 1  1 OP N1  Q 1  2 Ex.12 The energy of an electron in the second and third Bohr orbits of the hydrogen atom is – 5.42 × 10–12 ergs and – 2.41 × 10– 12 erg respectively. Calculate the wavelength of the emitted radiation when the electron drops from third to second orbit Sol. Here, h = 6.62 × 10– 27 erg E 3 = – 2.41 × 10– 12 erg E 2 = – 5.42 × 10– 12 erg E = E3 – E2 = – 2.41 × 10– 12 + 5.42 × 10– 12 Now we know that, E = hv 36 = Ex.11 The shortest wave length in H spectrum of Lyman series when RH = 109678 cm–1 is (A) 1215.67 Å (B) 911.7 Å (C) 1002.7 Å (D) 1127.30 Å (Ans B) Sol. For Lyman series n1 = 1 For shortest ‘l’ of Lyman sereis the energy differnece in two levels showing transition should be maximum (i.e. n2 = ). 3 4  = Since, E 3.01  10 12 = h  6.62  10 27 6.62  10 27  3  10 8 3.01  10 12  = 6.6 × 10–5 cm 1Å = 10–8 cm  = 6.6 × 103 Å Ex.13 Find the number of quanta of radiations of frequency 4.75 × 1013 sec–1, required to melt 100 g of ice. The energy required to melt 1 g of ice is 350 J Sol. E = nhv = n× 6.62× 10– 34 J sec × 4.75 × 1013 sec–1 = n × 31.445 × 10– 21 J Energy required to melt 100 g ice = 350 J × 100 = 35000 J n × 31.445 × 10– 21 = 35000 35000 n= = 1113 × 1021 31445 .  10 21 Ex.14 Calculate the number of photons emitted in 10 hours by a 60 W sodium lamp (of photon = 5893 Å) Sol. Energy emitted by sodium lamp in one sec. = Watt. × sec = 60 × 1 J Energy of photon emitted = 1 . hc  ATOMIC STRUCTURE =  6.625  10  34  3  10 8 5893  10 10 = 3.37 × 10– 19 J No. of photons emitted per sec. = 60 3.37  10 19  No. of photons emitted in 10 hours = 17.8 × 1019 × 10 × 60 × 60 = 6.41 × 1024 Ex.18 Show that the wavelength of a 150 g rubber ball moving with a velocity 50 m sec–1 is short enough to be observed Sol. Ex.15 Calculate the wavelength of a moving electron having 4.55 × 10– 25 J of kinetic energy Sol. Kinetic energy = (½mu2) = 4.55 × 10–23 J  u2 = 2  4.255  10 Ex.16 The minimum energy required to overcome the attractive forces electron and surface of Ag metal is 7.52 × 10– 19 J. What will be the maximum K.E. of electron ejected out from Ag which is being exposed to U.V. light of  = 360 Å – (A) 36.38 × 10– 19 Joule (B) 6.92 × 10–19 Joule (C) 57.68 × 10– 19 Joule (D) 67.68 × 10– 19 Joule (Ans B) hc Sol. Energy absorbed =  6.625  10 27  3.0  10 10 360  10  8 5.52 × 10–11 erg 5.52 × 10–18 Joule ( 7.52 × 10–19) – ( . 552 × 10–19) 6.92 × 10–19 Joule Ex.17 In hydrogen atom , an electron in its normal state absorbs two times of the energy as if requires to escape (13.6 eV) from the atom. The wave length of the emitted electron will be (A) 1.34 × 10–10 m (B) 2.34 × 10–10 m (C) 3.34 × 10–10 m (D) 4.44 × 10–10 m (Ans C) Sol. Energy absorbed by an atom = 2 × 13.6 = 27.2 eV Energy consumed in escape = 13.6 eV Energy converted into K.E. = 13.6 × 1.6 × 10–19 J v 2KE  m 2(13.6  16  10 19 ) 9.1  10 31 ms–1 = 2.18 × 106 ms–1  h mv = 6.63  10 34 9.1  10 31  2.1  10 6  = 6.625  10 27 150  50  10 2 = 8.83 × 10– 33 cm The wavelength is much longer than the l of visible region and thus it will not be visible. 9.108  10 31 6.625  10 34 h   = = 7.27 × 10– 7 meter mu 9.108  10 31  10 3 = = = =   25  u = 103 m sec–1 = h mu Given u = 50 m sec–1 = 50 × 102 cm sec– ; m = 150 g  = 3.34 × 10–10 m Ex.19 If an electron is present in n = 6 level. How many spectral lines would be observed in case of H atom– (A) 10 (B) 15 (C) 20 (D) 25 (Ans B) n(n  1) Sol. The no. of spectral lines is given by 2 when n = 6 then, the no. of spectral lines 6 x (6  1) 6 x 5   15 = 2 2 Ex.20 An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54 Å 1 mu 2  eV 2 Sol. We know that and =  1 h2 m  2 2  eV 2 m  or 1 h2 1 h2 V = 2m  2 2 = 2  m   e m 2  e h2 h h or u = or u2 = 2 2 mu m m  Substituting the values, we get 1 V = 2 (6.621034 )2 9.1081031 (1.541010 )2 1.6021019 = 63.3 volt Ex.21 What designation will you assign to an orbital having following quantum number – (a) n = 3, l = 1, m = –1 (b) n = 4, l = 2, m = +2 (c) n = 5, l = 0, m = 0 (d) n = 2, l = 1, m = 0 Sol. (a) Since l = 1 corresponds to p-orbital and m = –1 shows orientation either in x or y axis, thus this orbital refers to 3px or 3py ATOMIC STRUCTURE (d) 4d xy or (c) 5s Sol. 4d x2  y2 10.01 = x (d) 2pz  Percentage of isotope with atomic wt. 11.01 = 100 – x Ex.22 How many electrons in a given atom can have the following quantum numbers – (a) n = 4, l = 1 (b) n = 2, l = 1, m = – 1, s = + ½ (c) n=3 (d) n = 4, l = 2, m = 0 Sol. (a) l = 1 refers to p - subshell which has three or Average atomic wt. = Therefore, total number of electrons are 6. l = 1 refers to p - subshell, m = – 1 refers to px or py orbital whereas, s = +½ indicate for only 1 m1x1  m2 x 2 x1  x 2 Average atomic wt. = 10.81 orbitals (px, py and pz) each having two electrons. (b) Let the percentage of isotope with atomic wt. x  10.01  (100  x)  1101 . 100 x  10.01  (100  x)  1101 . x = 20 100 =  % of isotope with atomic wt. 10.01 = 20 % of isotope with atomic wt. 11.01 = 100 - x = 80 electron. Ex.25 From the following list of atoms, choose the (c) For n = 3, l = 0, 1, isotopes, isobars and isotones - 2 l=0 m=0 2 electrons l=1 m = –1 6 electrons l=2 m = –2 , –1, 0, +1, +2 16 8O Sol. 10 electrons Total electrons 18 electrons  (b) n = 2, (c) n = 2, = 2, m = – 1, s=0 = 2, m = +1, s=–½ = 2, m = + 1, s=–½ = 2. m = – 2, s=+½ (d) n = 3, Sol. (a) This set of quantum number is not permitted as value of ‘s’ cannot be zero. (b) This set of quantum number is not permitted as the value of ‘l ’ cannot be equal to ‘n’. (c) This set of quantum number is not permitted as the value of ‘l ’ cannot be equal to ‘n’. (d) This set of quantum number is permitted. Ex.24 Naturally occuring boron consists of two isotops whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron- , 14 6C , 40 20 Ca 238 92 U , Isotopes : 18 8O 40 235 238 ), ( 39 19 K , 19 K ), ( 92 U , 92 U ) 40 20 Ca ( 39 19 K , ) , ( 147 N , 40 20 Ca 14 6C ), ( 146 C , ) 16 8O ) Ex.26 Atomic radius is the order of 10– 8 cm. and Sol. Volume of nucleus = (4/3)pr3 = (4/3)p × (10–13)3 cm3 numbers are not permitted - l l l l 14 18 7N , 8 O , what fraction of atom is occupied by nucleus - Ex.23 Which of the following set of quantum n = 3, 40 19 K , nuclear radius is the order of 10–13 cm. Calculate Number of electrons are 2. (a) 235 92 U Isotones : 2n2 i.e. 2 × 32 = 18 electrons orbital , Isobars : ( 40 19 K , level is given by l = 2 means d-subshell and m = 0 refer to dz2 39 19 K ( 168 O , Alternatively, number of electrons for any energy (d) , volume of atom  = 4/3 pr3 = (4/3) p × (10– 8)3 cm3 Vnucleus 10 39   10 15 Vatom 10 24 or Vnucleus = 10– 15 × Vatom Ex.27 Nitrogen atom has Atomic number 7 & oxygen has Atomic number 8. Calculate the total number of electrons in nitrate ion Sol. No. of electrons in NO3– = (Electrons in N) + (3 × electrons in O) + [1(due to negative charge)] = 7 + 3 × 8 + 1 = 32