n = 3 ; E 3 = ev, n = 4 ; E 4 = ev, and so on. Noe, the energy of incident photon is E = h = = J, = ev, = ev

Transcription

1 Q 1: Hydrogen atom in its ground state excited by means of monochromatic radiation of wavelength 975 Å. How many different lines are possible in the resulting spectrum? You may assume the ionization energy for hydrogen atom as 13.6 ev: (a) 3 (b) 4 (c) 5 (d) Sol: (b) E n = - = - ev, For, n = 1 ; E 1 = ev, n = ; E = ev n = 3 ; E 3 = ev, n = 4 ; E 4 = ev, and so on. Noe, the energy of incident photon is E = h = = J, = ev, = 1.75 ev When hydrogen atom absorbs this incident photon, let the electron of ground state occupies the n th excited state. Then, E = - - ( ) = Rhc [1 - ] = 1.75 ev 13.6 (1 - ) = 1.75, 1 - =, 1/n = 1 = 0.065, n = 16 n = 4 Q : In question n. 1, calculate the longest wavelength amongst them. (a) 6563 Å (b) 809 Å (c) 1666 Å (d) Å Sol: (d) As shown in the energy level diagram, the longest wavelength corresponds to the transition 4 3 E 4 E 3 = h min, (- 1.51) = hc/ max, max =, = Å Q 3: The energy of an electron in excited hydrogen atom is -3.4 ev. The angular momentum of the electron according to the Bohr s theory is: (a) h/π (b) h/π (c) 3h/π (d) h/π Sol: (b) E n = - = - ev = - 3.4, n = = 4, or n = According to Bohr s theory, Angular momentum = = = Q 4: A single electron orbits around a stationary nucleus of charge Ze, where Z is a constant and e is the magnitude electron charge. It requires 47. ev to excite the electron from the second Bohr orbit to the third Bohr orbit. The value of Z is: (Ionization energy of H atom = 13.6 ev) (a) (b) 3 (c) 5 (d) 6 Sol: (c) E n = - Z /Rhc/n, E = E 3 E = - Z Rhc [ - ], = Z Rhc = 47. ev, Given Rhc = 13.6 ev This is the ionization energy of H-atom. Z (13.6) = 47., Z = = 5 Z = 5 Q 5: In q. no. 4, find the energy required to excite the electron from the third to the fourth orbit: (a) 0.03 ev (b) ev (c) 1.13 ev (d) ev Sol: (b) The energy required to excited the atom from third to fourth orbit is given by E = E 4 E 3 = - Z Rhc [ - ], = X (5) X 13.6 ev, = ev Q 6: In Q. no. 4, the wavelength of electromagnetic radiation required to move the electron from the first Bohr orbit to infinity is approximately: (a) 36 Å (b) 7 Å (c) 108 Å (d) 144 Å Sol: (a) The energy required to remove the electron from first orbit to infinity is equal to the ionization energy of the atom given by E - E 1 = - Z Rhc [ - ] = Z Rhc = (5) X 13.6 ev The wavelength corresponding to this energy (say) E - E 1 = h = hc/ = (5) X 13.6 ev = (5) X 13.6 X 1.6 X J, =, = m, = 36 X m, = 36 Å Q 7: In Q. no.4, the ratio of the kinetic energy to the potential energy of the electron in the first Bohr orbit is: (a) -1 (b) (c) 1/ (d) 1/4 Sol: (c) Kinetic energy of electron in the first Bohr orbit K 1 = - E 1 = Z Rhc = (5) X 13.6 ev, = 340 ev Potential energy of electron in the first Bohr orbit U 1 = E 1 = ev, or K 1 /U 1 = -

2 Q 8: The ionization energy of hydrogen like a Bohr atom is 4 Rydbergs. The Atomic no Z of the atom is: (a) 1 (b) (c) 4 (d) > 4 Sol: (b) E n = - Z Rhc/n, where Rhc = 1 Ryberg (given), The ionization energy of the given atom is given by E - E 1 = Z Rhc = 4 Rydbergs, or Z = 4, or Z = Q 9: In Q. no. 8, the energy required to excite the electron from the first Bohr orbit to the second Bohr orbit is: (a) 1.0 Rydbergs (b) 1.6 Rydbergs (c) 3.0 Rydbergs (d).6 Rydbergs Sol: (c) The energy required to excite the electron from n = 1 to n = is given by E E 1 = - Z Rhc [ - ], = Z Rhc = X 4 X 1 Rydbergs, = 3 Rydbergs. Q 10: In Q. no.8, what is the wavelength of radiation emitted when the electron jumps from first excited state to the ground state? (a) Å (b) Å (c) 1515 Å (d) 3034 Å Sol: (b) In problem no. 9, if is the wavelength corresponding to the energy transition of 3 Rydbergs then hc/ = 3 Rydbergs, or =, 1 Rydbergs = Rhc = = =, 303 X m, or Å Q 11: Electrons in hydrogen like atoms (Z = 3) make transitions from the fifth to fourth orbit and front the fourth to third orbit. The resulting radiations are incident normally on a metal plate and eject photo electrons. The stopping potential for the photoelectrons ejected by shorter wavelength is 3.95 V. The stopping potential for the photoelectrons ejected by longer wavelength is:[rydberg s constant R = m -1 ; h = J-s] (a) 0.95 V (b) 0.75 V (c) 0.55 V (d) 0.35 V Sol: (b) The wave number, according to Bohr s theory is given by = 1/ = Z R ( - ) For transition 5 4, 1/ 1 = Z R ( - ) = (3) X X 10 7 X, 1 = X 10 7 m For transition 4 3, 1/ = Z R ( - ) = (3) X X 10 7 X, =.089 X 10 7 m The stopping potential for the shorter wavelength is 3.95 V (given) hc/ = + ev = - ev, = X X 3.95, = 3. X J, = ev = ev For the longer wavelength 1, hc/ 1 = + ev 1 ev 1 = -, = - 3. X 10-19, ev 1 = 1.06 X J, V 1 = = 0.75 ev. Q 1: In Q. no. 11, calculate the work function of the metal: (a) 1 ev (b) ev (c) 3 ev (d) 4 ev Sol: (b) The wave number, according to Bohr s theory is given by = 1/ = Z R ( - ) For transition 5 4, 1/ 1 = Z R ( - ) = (3) X X 10 7 X, 1 = X 10 7 m For transition 4 3, 1/ = Z R ( - ) = (3) X X 10 7 X, =.089 X 10 7 m The stopping potential for the shorter wavelength is 3.95 V (given) hc/ = + ev = - ev, = X X 3.95, = 3. X J, = ev = ev For the longer wavelength 1, hc/ 1 = + ev 1 ev 1 = -, = - 3. X 10-19, ev 1 = 1.06 X J, V 1 = = 0.75 ev. Q 13: when photons of energy 4.5 ev strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy T A ev and de Broglie wavelength λ A. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 ev is (K B = K A ) ev. If the de Broglie wavelength of these photoelectrons is λ B = λ A, then:

3 (a) The work function of A is.5 ev (b) The work function of B is 4.0 ev (c) K A =.00 ev (d) K B =.75 ev Sol: (a,b,c) From Einstein s photoelectric equation, h = + (E k ) max For metal A 4.5 ev = A + k A,.(i), For metal B 4.70 ev = B + k B.(ii) The de Brogile wavelength = 1/, = = (given) k A = 4k B, but k B = (k A 1.50) ev (given), k A /4 = k A 1.50, k A = 1.50 k A = ev and k B = 0.50 ev, Putting these values in Eqs. (i) and (ii), we get, a =.5 ev and B = 4.0 ev Q 14: A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.0 ev and ev respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.5 ev and 5.95 ev respectively. If the ionization energy of H atom is 13.6 ev, then the value of Z is: (a) 6 (b) 3 (c) 4 (d) 5 Sol: (b) In the first case, the total energy. Q 15: In Q. no 114, the value of n is: (a) 6 (b) 3 (c) 4 (d) 5 Sol: (a) Radiated during transition from n th state to 1 excited (n = ) state is given by = 10.0 = = 7.0 ev From Bohr s theory, E n E 1 = 13.6 Z [ - ] = 7.0 ev (i) For the second case E n E 3 = 13.6 Z [ - ], = = 10.0 ev Solving Eqs. (i) and (ii), we get, Z = 3 and n = 6 Q 16: The wavelength of the first line of Lyman series for Hydrogen is identical to that of the second line of Balmer series for some hydrogen like ion X. The energy of second and fourth level is: (Given: Ground state binding energy of hydrogen atom = 13.6 ev) (a) 17.6 ev; ev (b) 13.6 ev ; ev (c) 13.6 ev;.18 ev (d) ev; ev Sol: (b) For the first Lyman line of H-atom 1/ 1 = R (1 - ) = 1/ = RZ ( - ) =, Given 1 =, or =, Given Rhc = 13.6 ev, E n = - = - ev, E n = - ev, Thus, E = - = ev, and E 4 = - = ev Q 17: In Q. no. 16, the ionization energy of the ion and its atomic number is: (a) 17.6 ev;3 (b) 54.4 ev; (c) 13.6 ev; 1 (d) none of these Sol: (b) Ionisation energy is E n = - ev, E - E 1 = [ - 1] = 54.4 ev Q 18: If elements with principle Quantum number n > 4 were not allowed in nature, the no possible element would be: (a) 60 (b) 3 (c) 4 (d) 64 Sol: (a) The total number of electron in an orbit of principal quantum number n is n. Hence, the required number of elements is equal to = [ ], = [ ] = X 30 = 60 Q 19: The splitting of a line spectrum into groups under the effect of electric and magnetic field is called: (a) Zeeman effect (b) Compton effect (c) Bohr effect (d) Heisenberg effect Sol: (a) Q 0: What will be the ratio of the de Broglie Wavelengths of a photon and an α-particle of the same energy? (a) :1 (b) 1: (c) 4:1 (d) 1:4 Sol: (a) =, = = = Q 1: The transition from state n = 4 to n = 3 in a hydrogen like atoms result in ultraviolet radiations. Infrared radiations will be obtained in the transition: (a) 1 (b) 3 (c) 4 (d) 5 4

4 Sol: (d) The wavelength of infrared radiation is greater than that of ultraviolet radiation. 1/ = RZ ( - ), Thus, - or, or a : b : c : d = : : :, = : : :, or maximum for 5 4 Q : An α particle of energy 5 MeV is scattered through by a fixed uranium nucleus. The distance of closest particleaaproach is of the order of: (a) 1 Å (b) cm (c) 10-1 cm (d) cm Sol: (c) r 0 =, Here Z = 9; E k = 5 MeV = 5 X 10 6 X 1.6 X J So, r 0 =, 10-1 cm Q 3: If 13.6 ev energy is required to ionize the hydrogen atom, then the energy required to remove from an electron from n = is: (a)10. ev (b) 0 (c) 3.4 ev (d) 6.8 ev Sol: (c) E n = -, E - E = ( - ) = = 3.4 ev Q 4: Which of the following atoms has the lowest ionization potatial? Sol: (b) (a) 14 N (b) 55 Cs (c) 18 Ar (d) O Cs has largest size among the four atom given. Thus, electrons in the outermost orbit is at largest 16 8 distance in Cs among the four atoms given and so the electrostatic force experienced by electrons due to nucleus will be minimum in Cs. Hence, the energy required to liberate electron from the outermost orbit will be lowest in Cs. Q 5: The wavelengths involved in the spectrum of deuterium ( 1 D ) are slightly different from that of hydrogen spectrum, because: (b) Nuclear forces are different in the two cases (a) Size of the two nuclei are different (c) Masses of the two nuclei are different (d) Attraction between the electron and the nucleus is different in the two cases Sol: (c) Wavelength of spectral line depends on the Rydberg s constant R = e 4 /8 0 ch 3 where the reduced mass = nm/m + M, m is the mass of electron and M is the mass of the nucleus of the atom. Since the masses of the two nuclei are different hence the wavelength are different. Q 6: 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 Lі + is: (a) 30.6 ev (b) 13.6 ev (c) 3.4 ev (d) 1.4 ev Sol: (a) E n = (Z/n) ev, For the first excited state, n = and for Li +, Z = 3 E = X = = ev, Thus, the ionization energy for the excited state of Li + is 30.6 ev. n = 4 Q 7: The diagram shows the energy levels for an electron in a certain atom. Which transition represents the emission of a photon with the most energy? n = 3 (a) ііі (c) і (b) іv (d) іі Sol: (a) E n = - Rhc [1/n ], E 4 E 3 = Rhc [ - ] = (Rhc).(II) E 4 E = Rhc [ - ] = (Rhc)..(IV), E E 1 = Rhc [1 - ] = (Rhc) (i) (ii) (iii).. (iv) (III) E 1 E 3 = Rhc [ - 1] = - (Rhc)..(I), Hence, the transition (III) gives the most energy. Q 8: If the kinetic energy of a free electron doubles, its de Broglie wavelength changes by a factor: (a) 1 (b) (c) 1 (d) n = n = 1

5 Sol: (c) =,, = =, = 1 / Q 9: A alpha nucleus of energy ½ mv bombards a heavy nuclear target of charge Ze. Then the distance of closet approach for the alpha nucleus will be proportional to: (a) v (b) 1/m (c) 1/v (d) 1/Ze Sol: (c) Since the target nucleus is heavy, it can be assumed that the target nucleus remains stationary when - particle approaches near it. The kinetic energy of the -particles converts into electrostatic potential energy at the distance of closet approach. Hence E k = m = This gives, r =, =, Thus, r ; r ; r Ze Q 30: The transition of an electron from n = 5, 6... to n 1 = 4 gives rise to: (a) P fund series (b) Lymann series (c) Brackett series (d) Paschen series Sol: (c) Q 31: The ground state energy of hydrogen atom is ev. What is the potential energy of the electron in this state? (a) 0 ev (b) 13.6 ev (c) -7. ev (d) ev Sol: (c) Total energy of electron in the n th state is given by E n = - = ev [For hydrogen atom Z = 1], Potential energy of electron in the n th state is given by, U n = - = E n = - 7. ev Q 3: The first member of Ballmer s series of hydrogen has a wavelength λ. The wavelength of second member is: (a) 7 0 Sol: (c) For Balmer s series, (b) 7 0 (c) 0 7 1/ = R [ - ] where n = 3, 4, 5 etc The wavelength first member is 1 = = = (given) (i) (d) none of these The wavelength of second member is = =, = X = Q 33: Energy E of a photon ejected when the electron jumps from n = 3 state to n = state of hydrogen is approximately: (a) 0.85 ev (b) 1.5 ev (c) 1.9 ev (d) 3.4 ev Sol: (c) E n = ( ) ev, or E 3 E = 13.6 [ - ] = = 1.89 ev, 1.9 ev Q 34: In hydrogen spectrum, the shortest wavelength in Balmer series is λ. The shortest wavelength in Brackett series will be: (a) λ (b) 4λ (c) 9λ (d) 16λ Sol: (b) Shortest wavelength of Balmer series, b = = = (given) and that of Bracket series, b = = = 4 Q 35: In a hypothetical Bohr hydrogen atom, the mass of the electron is doubled. The energy E 0 and the radius r 0 of the first orbit will be: (a) E 0 = - 7. ev; r 0 = a 0 (b) E 0 = ev; r 0 = a 0 / (c) E 0 = - 7. ev; r 0 = a 0 / (d) E 0 = ev; r 0 = a 0 Sol: (c) Radius of the electron orbit, r = (for hydrogen atom Z = 1), So, r If m is doubled, the value of r is reduced to r/. Also Energy of electron in n th orbit, E = -, So, E m, If m is doubled, the energy because E. Since, r 0 = a 0 and E 0 = ev, radius becomes a 0 / and energy becomes, Therefore, when mass of the electron is doubled, the E 0 = - 7. ev

6 Q 36: The angular momentum of an electron in a hydrogen atom is proportional to: (a) 1/ 3 (b) 1/r (c) 3 (d) r Sol: (c) Angular momentum L = m r, Here =, i.e.., and r =, i.e.., r n, or L n or L r 1/ Q 37: The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition: (a) 1 (b) 3 (c) 4 (d) 5 4 Sol: (d) Larger wavelength of paschen series [n 3] where n = 4, 5, 6, etc. and all the lines of Brackett series [n 4] where n = 5, 6, 7, etc. lie in the infrared region. Therefore transition from n = 5 to n = 4 is the only correct choice in the given four options. Q 38: For explanation of this theory, Bohr used which of the following principles: (a) Conservation of energy (b) Conservation of linear momentum (c) Conservation of angular momentum (d) Conservation of moment of inertia Sol: (c) Q 39: Energy of an electron in n th of hydrogen atom is: (a) K m 4 e (b) 4 mke (c) n h n h n h (d) Km 4 e n h 4 e mk Sol: (a) E n = -, Taking Z = 1 (for hydrogen atom), E n = - Also, K = 0 =, or E n = - Q 40: The energy of an electron in excited hydrogen atom is -3.4 ev. Then according to Bohr s theory the angular momentum of the electron in J-s is: (a) (b) (c) (d) (e) Sol: (d) e n = - = - 3.4, or n =, Angular momentum, L n = = =, = =.11 X J-s Q 41: In the Bohr model of hydrogen atom, let R, v and E represent radius of the orbit, speed of electron and total energy of the electron respectively. Which one of the following quantities is proportional to the quantum no n? (a) R/E (b) E/v (c) RE (d) vr Sol: (d) Angular momentum L = m r = nh/, or r n Q 4: Consider a special line resulting from the transition n = 5 to n = 1 in the atoms and ions given below. The shortest wavelength is produced by: (a) Helium atom (b) Deuterium atom (c) Singly ionized helium (d) doubly ionized lithium (e) Ten times ionized sodium atom Sol: (e) The wavelength of spectral line for energy transition from n = 5 orbit to n = 1 orbit is given by =, So, Q 43: For an electron in the second orbit of Bohr hydrogen atom, the moment of linear momentum is: (a) ph (b) ph (c) h/π (d) h/π Sol: (c) The moment of linear momentum is called the angular momentum L which is for n th orbit given by, L = nh/, for n =, L = h/ Q 44: When the wave the hydrogen atom comes from infinity into the first orbit, then the value of wave number is: (a) cm -1 (b) 1097 cm -1 (c) 109 cm -1 (d) None of these Sol: (a) = = R [1 - ] = R, = X 10 7 m -1, = X 10 5 cm -1, = cm -1 Q 45: If the series limit wavelength of Lyman series for hydrogen atom is 91 Å, then the series wavelength limit for the Balmer series for the hydrogen atom is: (a) 91 Å (b) 91 Å (c) 91 4 Å (d) 91/ Å

7 Sol: (c) The series wavelength limit of Lyman series is given by, 1 = = and that of Balmer series is given by, b = = = 4 1, = 4 X 91 Å Q 46: The angular speed of the electron in the n th orbit of Bohr s hydrogen atom is: (a) Inversely proportional to n (b) Inversely proportional to n (c) Inversely proportional to n 3 (d) Directly proportional to n Sol: (c) We know that and r n, Since, = /r, so 1/n 3 Q 47: The color of second line of Balmer series is: (a) Blue (b) Yellow (c) Red (d) Violet Sol: (a) The wavelength of second line of Balmer series is, = =, = = X 10-7 m, = 4860 Å Q 48: Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 ev. The energy needed to remove the electron from the ion in ground state is: (a) 40.8 ev (b) 54.4 ev (c) 7. ev (d) 13.6 ev Sol: (b) The energy level of ground state is n = 1 and for its first excited slate is n = let the energy state is E 1 and that in its first excited state is E. Then the excitation energy of a hydrogen like ion, in its first excitation state is given by E = E 1 ( - ) = 40.8 ev (given) or E 1 [ - ] = 40.8, E 1 = = 54.4 ev Q 49: If element with principle quantum no. n > 4 were not allowed in nature, the no. of possible elements would have been: (a) 4 (b) 3 (c) 60 (d) 64 Sol: (c) For n = 4, l = 0, 1,, 3, Now the number of electrons in subshells corresponding to l = 0, 1,, 3 are, 9, 18 and 3 respectively. Hence the total number of possible elements would be = 60 Q 50: A photon of energy 10. ev collide in elastically with hydrogen atom in ground state. After few micro seconds another photon of energy 15 ev collides with in elastically with same hydrogen atom. Finally by suitable detector, we find: (a) Photon of energy 10 ev and electron of energy 1.4 ev (b) Photon of energy 3.4 ev and electron of energy 1.4 ev (c) Two photon of energy 3.4 ev (d) Two photon of energy 10. ev Sol: (a) In the first inelastic collision of photon with the hydrogen atom, the atom gets excited after capturing energy of 10. ev which is less than the ionistaion energy (13.6 ev) of the H-atom. Since the electron in excited stated can remain only for maximum 10-9 s so it will soon (almost immediately) return to its ground state and this results an emission of first line of Lyman state. The energy of emitted photon is E E 1 = 13.6 [ - ] ev = 10. ev Now another photo of energy 15 ev strikes with the H-atom again inelastically. This energy is greater than the ionization energy of H-atom. Hence the electron is removed and the atom gets ionized. The excess energy with which electron comes out is ( ) = 1.4 ev. Hence, we finally detect a photon of energy 10. ev and an electron of energy 1.4 ev

8 Answers 1. b. d 3. b 4. c 5. b 6. a 7. c 8. b 9. c 10. b 11. b 1. b 13. b,c 14. b 15. a 16. b 17. b 18. a 19. a 0. a 1. d. c 3. c 4. b 5. c 6. a 7. a 8. c 9. c 30. c 31. c 3. c 33. c 34. b 35. c 36. c 37. d 38. c 39. a 40. d 41.d 4. e 43. c 44.a 45. c 46. c 47. a 48. b 49.c 50. a