Chapter 3 Atomic Structure Answers

Atomic Structure 11 CHAPTER Atomic Structure 1. Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is (a) 5 eV (b) 4 eV (c) 5.5 eV (d) 4.5 eV 2. The de-Broglie’s wavelength of electron present in first Bohr orbit of ‘H’ atom is 0.529 (a) Å (b) 2 × 0.529 Å 2 (c) 0.529 Å (d) 4 × 0.529 Å 3. 4. Which of the following statements is false? (a) Photon has momentum as well as wavelength. (b) Splitting of spectral lines in electrical field is called Stark effect. (c) Frequency of emitted radiation from a black body goes from a lower wavelength to higher wavelength as the temperature increases. (d) Rydberg constant has unit of energy. The radius of the second Bohr orbit for hydrogen atom is (Planck’s constant (h) = 6.6262 × 10–34 J s; mass of electron = 9.1091 × 10–31 kg; charge of electron = 1.60210 × 10–19 C; permittivity of vacuum (0) = 8.854185 × 10–12 kg–1 m–3 A2) (a) 0.529 Å (c) 1.65 Å 5. 6. (b) 2.12 Å (d) 4.76 Å If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of He+ is 5A 36 A 9A 36 A (a) (b) (c) (d) 9 5 5 7 The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with (a) Paschen series (b) Brackett series (c) Lyman series (d) Balmer series. 7. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron respectively, then the value of h/ (where  is wavelength associated with electron wave) is given by (a) meV (b) 2meV (c) meV (d) 2meV 8. The total number of orbitals associated with the principal quantum number 5 is (a) 20 (b) 25 (c) 10 (d) 5 9. Which of the following is the energy of a possible excited state of hydrogen? (a) –3.4 eV (b) +6.8 eV (c) +13.6 eV (d) –6.8 eV 10. If the principal quantum number n = 6, the correct sequence of filling of electrons will be (a) ns  np  (n – 1)d  (n – 2)f (b) ns  (n – 2)f  (n – 1)d  np (c) ns  (n – 1)d  (n – 2)f  np (d) ns  (n – 2)f  np  (n – 1) 11. At temperature T, the average kinetic energy of any particle is (a) (b) (c) (d) 3 kT. The de Broglie wavelength follows the order 2 thermal proton > visible photon > thermal electron thermal proton > thermal electron > visible photon visible photon > thermal electron > thermal neutron visible photon > thermal neutron > thermal electron. 12. The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is (a) 5, 0, 1,  1 2 (b) 5, 0, 0,  1 2 (c) 5, 1, 0,  1 2 (d) 5, 1, 1,  1 2  Z2  13. Energy of an electron is given by E  2.178  1018 J  2  . n  Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be (h = 6.62 × 10 –34 J s and c = 3.0 × 108 m s–1) (a) 8.500 × 10–7 m (b) 1.214 × 10–7 m (c) 2.816 × 10–7 m (d) 6.500 × 10–7 m 14. The electrons identified by quantum numbers n and l: (1) n = 4, l = 1 (2) n = 4, l = 0 (3) n = 3, l = 2 (4) n = 3, l = 1 can be placed in order of increasing energy as (a) (4) < (2) < (3) < (1) (b) (2) < (4) < (1) < (3) (c) (1) < (3) < (2) < (4) (d) (3) < (4) < (2) < (1) 15. A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at (a) 1035 nm (b) 325 nm (c) 743 nm (d) 518 nm 16. The energy required to break one mole of Cl—Cl bonds in Cl 2 is 242 kJ mol–1. The longest wavelength of light capable of breaking a single Cl—Cl bond is (c = 3 × 10 8 m s –1 and NA = 6.02 × 10 23 mol –1) (a) 494 nm (b) 594 nm (c) 640 nm (d) 700 nm 17. Ionisation energy of He+ is 19.6 × 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is (a) 8.82 × 10–17 J atom–1 (b) 4.41 × 10–16 J atom–1 –17 –1 (c) – 4.41 × 10 J atom (d) – 2.2 × 10–15 J atom–1 18. Calculate the wavelength (in nanometre) associated with a proton moving at 1.0 × 103 m s–1. (Mass of proton = 1.67 × 10–27 kg and h = 6.63 × 10 –34 J s) (a) 0.032 nm (b) 0.40 nm (c) 2.5 nm (d) 14.0 nm 19. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainty with which the position of the electron can be located is (h = 6.6 × 10–34 kg m2 s–1, mass of electron, em = 9.1 × 10–31 kg) (a) 1.52 × 10–4 m (b) 5.10 × 10–3 m –3 (c) 1.92 × 10 m (d) 3.84 × 10–3 m 20. The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol–1. The energy required to excite the electron in the atom from n = 1 to n = 2 is (a) 9.84 × 105 J mol–1 (b) 8.51 × 105 J mol–1 5 –1 (c) 6.56 × 10 J mol (d) 7.56 × 105 J mol–1 1 2 1 (d) n = 4, l = 0, m = 0, s =  2 (c) n = 3, l = 2, m = 1, s =  22. Uncertainty in the position of an electron (mass = 9.1 × 10–31 kg) moving with a velocity 300 m s–1, accurate upto 0.001% will be (h = 6.6 × 10–34 J s) (a) 19.2 × 10–2 m (b) 5.76 × 10–2 m –2 (c) 1.92 × 10 m (d) 3.84 × 10–2 m 23. According to Bohr’s theory, the angular momentum of an electron in 5th orbit is h h h h (a) 25 (b) 1.0 (c) 10 (d) 2.5     24. Which of the following statements in relation to the hydrogen atom is correct? (a) 3s orbital is lower in energy than 3p orbital. (b) 3p orbital is lower in energy than 3d orbital. (c) 3s and 3p orbitals are of lower energy than 3d orbital. (d) 3s, 3p and 3d orbitals all have the same energy. 25. In a multi-electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic and electric fields? (i) n = 1, l = 0, m = 0 (ii) n = 2, l = 0, m = 0 (iii) n = 2, l = 1, m = 1 (iv) n = 3, l = 2, m = 1 (v) n = 3, l = 2, m = 0 (a) (i) and (ii) (b) (ii) and (iii) (c) (iii) and (iv) (d) (iv) and (v) 26. The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097 × 107 m–1) (a) 91 nm (b) 192 nm (c) 406 nm (d) 9.1 × 10–8 nm 27. Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively (a) 12 and 4 (b) 12 and 5 (c) 16 and 4 (d) 16 and 5 28. Which of the following sets of quantum numbers is correct for an electron in 4f orbital? 1 2 1 (b) n = 4, l = 4, m = –4, s =  2 1 (c) n = 4, l = 3, m = +1, s =  2 1 (d) n = 3, l = 2, m = –2, s =  2 (a) n = 4, l = 3, m = +4, s =  21. Which of the following sets of quantum numbers represents the highest energy of an atom? (a) n = 3, l = 0, m = 0, s =  1 2 (b) n = 3, l = 1, m = 1, s =  1 2 29. The orbital angular momentum for an electron revolving in h l ( l  1)  2 an orbit is given by s-electron will be given by (a)  1 h  2 2 (b) zero inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen? (a) 3  2 (b) 5  2 (c) 4  1 (d) 2  5 . This momentum for an (c) h 2 (d) 2 h 2 30. The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second is approximately (Planck's constant, h = 6.63 × 10–34 J s) (a) 10–33 metres (b) 10–31 metres –16 (c) 10 metres (d) 10–25 metres. 31. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following 32. Uncertainty in position of a minute particle of mass 25 g in space is 10 –5 m. What is the uncertainty in its velocity (in m s–1)? (h = 6.6 × 10–34 J s) (a) 2.1 × 10–34 (b) 0.5 × 10–34 (c) 2.1 × 10–28 (d) 0.5 × 10–23 33. In a hydrogen atom, if energy of an electron in ground state is 13.6 eV, then that in the 2nd excited state is (a) 1.51 eV (b) 3.4 eV (c) 6.04 eV (d) 13.6 eV ANSWER KEY 1. (d) 13. (b) 25. (d) 2. (b) 14. (a) 26. (a) 3. (c) 15. (c) 27. (b) 4. (b) 16. (a) 28. (c) 5. (c) 17. (c) 29. (b) 6. (d) 18. (b) 30. (a) 7. (d) 19. (c) 31. (b) 8. (b) 20. (a) 32. (c) 9. (a) 21. (c) 33. (a) 10. (b) 22. (c) 11. (c) 23. (d) 12. (b) 24. (d) 1. (d) : K.E. = h – h0 = E – W0 where, K.E. = Kinetic energy of ejected electron = Stopping potential E = Energy absorbed, W0 = Work function hc 6.626  1034  3  108  E = h = = 7.9512 × 10–19 J = 4.96 eV  250  109 Then, 0.5 = 4.96 + W0 W0 = 4.46  4.5 eV (b) : rn = 0.529 (n)2 Å nh nh nh mvr   mv   2 2 r 2  (0.529) h h     2  0.529 = 2 × 0.529 Å mv nh 3. (c) : When a black body is heated, more and more energy is absorbed by its atoms and they emit radiations of higher and higher frequency, i.e., black body emits radiation from higher wavelength to lower wavelength. n2a0 Å 4. (b) : Radius of nth orbit for H-atom is r  Z 2 (2)  0.529 r Å [  n = 2, for second orbit] 1 r = 2.12 Å 5. (c) : The shortest wavelength of hydrogen atom in Lyman series is from n1 = 1 to n2 =   1 1 1   RZ 2  2  2    n1 n2  2. 1 1 1  1  R 2  2   R {  Z = 1, for hydrogen} 1   1 A 1  R A The longest wavelength in Paschen series of He+ is from n1 = 3 to n2 = 4 1 7 7 1 1  1 2 1 1  4  RZ 2    2  (2)  2  2   2  3 4  A 16  9 36 A 2 A  n1 n2  36 A 7 6. (d) : r = 211.6 pm = 2.11 Å n2 = 2.11 Å (Z = 1) r = 0.529 × Z 2  n =4n=2 In Balmer series, transition of electron occurs from higher orbitals to orbital of n = 2.  7. 2 = (d) :  h  2meV  h  p h  2m K . E. h 2 meV ( K.E. = eV] (b) : Number of orbitals in nth shell = n2 = (5)2 = 25 (a) : Energy of electron in the nth orbit of H-atom is, 8. 9. En  13.6 2 n eV= 13.6 22  3.4 eV 10. (b) : The electrons are filled as per (n + l) rule. Orbital having lower (n + l) value is filled first and when (n + l) values are same, the one having lower n value is filled first. Hence, the sequence of filling electrons in sixth period will be 6s – 4f – 5d – 6p i.e. (ns)  (n – 2)f  (n – 1)d  np 11. (c) :   h 2mE ; where, m is mass and E is kinetic energy of the particle. At constant temperature, E is constant.  1 m 12. (b) : Rb (Z = 37) : [Kr] 5s1 1 n = 5, l = 0, m = 0, s =  2 13. (b) : E = E E E 2.178 10 2.178 10 18 1 1 18 Z2 1 1 n12 n22 (2)2 (1)2 3 2.178 10 18 1.6335 10 18 J 4 hc 6.62 10 34 J s 3 108 m hc  E 1.6335 10 18 J  = 12.14 × 10–8 m or  = 1.214 × 10 –7 m 14. (a) : (1) n = 4, l = 1  4p (2) n = 4, l = 0  4s (3) n = 3, l = 2  3d (4) n = 3, l = 1  3p Increasing order of energy is 3p < 4s < 3d < 4p (4) < (2) < (3) < (1) Alternatively, for (1) n+l=5; n=4 (2) n+l=4; n=4 (3) n+l=5; n=3 (4) n+l=4; n=3 Lower n + l means less energy and if for two subshells n + l is same than lower n, lower will be the energy. Thus correct order is (4) < (2) < (3) < (1). 15. (c) : We know that, E = h = hc/ hc hc E = E1 + E2 or hc    1  2 1 1 1 1 1 1        1  2 355 680  2   2 355  680  742.769 nm  743 nm 680  355 16. (a) : Energy required to break 1 mol of bonds = 242 kJ mol–1  Energy required to break 1 bond  242  10 3 6.02  10 23 J hc ; Given, c = 3 × 108 m s–1  6.63  1034  3  108   We know that, E   242  10 3    6.02  10 23 6.63  1034  3  108  6.02  1023 242  10 3 = 0.494 × 10–6 m = 494 nm 17. (c) : I.E.(He+) = 19.6 × 10–18 J atom–1 E1 (for H) × Z2 = I.E. E1 × 4 = –19.6 × 10–18 E1 (for Li2+) = E1 (for H) × 9 19.6  1018  9  4 25. (d) : Orbitals having same (n + l) value in the absence of electric and magnetic field will have same energy. – 4.41 × 10–17 J atom–1 18. (b) : According to de-Broglie’s equation, Given, v = 1.0 × 103 m s –1 6.63 10 34 1.67 10 27 1.0 103 3.9 10 10 22. (c) : According to Heisenberg’s uncertainty principle, h x  p  4 h h x  (m  v)    x  4 4m  v 0.001 300 3 10 3 m s 1 Here v 100 6.63  1034  x  1.92 10 2 m 4  3.14  9.1 1031  3  103 nh 23. (d) : Angular momentum of the electron, mvr  2 when n = 5 (given) 5h h 2.5  Angular momentum = 2 24. (d) : For hydrogen the energy order of orbital is 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f h mv m  0.4 nm 19. (c) : Given, velocity of e–, v = 600 m s–1 Accuracy of velocity = 0.005% 600 0.005 0.03 100 According to Heisenberg’s uncertainty principle, 6.6 10 34 h x x m v = 1.92 × 10–3 m 4 4 3.14 9.1 10 31 0.03 20. (a) : The ionisation of H-atom is the energy absorbed when the electron in an atom gets excited from first shell (E1) to infinity (i.e., E) I.E = E – E1 1.312 × 10 6 = 0 – E1 E1 = –1.312 × 106 J mol–1   26. (a) : 1 R  1  1  1.097 107 m n 2 n 2   1 2    = 91 × 10–9 m = 91 nm 6 E = E2 – E1  1.312  10  ( 1.312  106 ) 4 6 6  1.312  10  5.248  10 = 9.84 ×10 5 J mol–1 4 21. (c) : n = 3, l = 0 represents 3s orbital n = 3, l = 1 represents 3p orbital n = 3, l = 2 represents 3d orbital n = 4, l = 0 represents 4s orbital The order of increasing energy of the orbitals is : 3s < 3p < 4s < 3d 1 12 27. (b) : 24Cr  1s2 2s2 2p6 3s2 3p6 3d5 4s1 we know for p, l = 1 and for d, l = 2. For l = 1, total number of electrons = 12 For l = 2, total number of electrons = 5 28. (c) : For 4 f orbital electrons, n = 4 v 6 6 E2   1.312 210   1.312  10 4 (2) Energy of electron in second orbit (n = 2)  Energy required when an electron makes transition from n = 1 to n = 2 1 1 2 [2p6 and 3p 6 ] [3d 5] s p d f l = 3 (because 0 1 2 3 ) m = +3, +2, +1, 0, –1, –2, –3 s = 1/2 29. (b) : The value of l (azimuthal quantum number) for s-electron is equal to zero. Orbital angular momentum = l (l  1)  h 2 Substituting the value of l for s-electron =  30. (a) :  34 0(0 1) h 2 0 h 6.63  10  1000 = 1.105 × 10–33 metres. mv 60  10 31. (b) : The electron has minimum energy in the first orbit and its energy increases as n increases. Here n represents number of orbit, i.e. 1 st, 2nd, 3 rd .... The third line from the red end corresponds to yellow region i.e. 5. In order to obtain less energy electron tends to come in 1st or 2 nd orbit. So jump may be involved either 5  1 or 5  2. Thus option (b) is correct here. 32. (c) : According to Heisenberg uncertainty principle, h x  m v  4 v 6.6 10 34 1000 4 3.14 25 10 5 2.1 10 28 ms 1 33. (a) : 2nd excited state will be the 3rd energy level. En  13.6 eV or E 13.6 1.51 eV 9 n2