Atomic Model and Hydrogen Atom 1. Ifthe binding energy of ground state electron in a hydrogen atom is 13.6 eV, then, the energy required to remove the electron from the second excited state of Li will be: x%10"'eV. The value of xis [31 Jan, 2023 (Shift-I1)] 2. Alightof energy 12.75 eV is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one ofits excited states. The angular momentum of the atom in the excited state is 2x10" eVs. The value of x x is (use h=4.14%10" 5 eVs,c=3x108 ms"), [1 Feb, 2023 (Shitt-1)] A 125 eV electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be: [12 April, 2023 Shitt-1] @2 1 ©3 @4 4. The angular momentum for the electron in Bohr’s orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum 3 will be: 110 April, 2023 Shite-t] @e (b) zero 2 OL (@ 2L ‘The energy levels ofan atom is shown in figure. Which one of these transitions will resul in the emission ofa photon of ‘wavelength 124.1 nm? Given (h= 6,62 x 10 Js) [25 Jan, 2023 (Shift-11)] 6. 8. 1—A_B__Pooey 2 | Ct 2 pee -10.0eV @B (A @c @D An electron of a hydrogen like atom, having Z=4, jumps from 4" energy state to 2 energy state, The energy released in this process, will be: (Given Reh =13.6eV) ‘Where R = Rydberg constant c= Speed of light in vacuum. A= Planck’s constant (@ 136er © 105 ev (© 34ev © 408 ev ‘The radius of electron’s second stationary orbit in Bohr's atom is R. The radius of 3rd orbit will be : [31 Jan, 2023 (Shift-11)] (a) RI (6) 2.25R © 3R (@ 9R A small particle of mass m moves in such a way that its U1 Feb, 2023 (Shift-I1)] ‘moo?r? where @ is constant and r is the distance ofthe particle from origin, Assuming Bohe's quantization of momentum and circular orbit, the radius of n* orbit will be proportional to, ‘16 April, 2023 (Shi «@ vn (in On wt a9. ‘The radius of 2" orbit of He'of Bors models, and that of fourth orbit of Be" is represented as r,. Now the ratio rir isx 1. The value of xis [13 April, 2023 (Shift 10. The radius of fifth orbit of the Li" is__ radius of hydrogen atom = 0.51A 16 April, 2023 (Shift-1)| 11, Jd, and fd, are the impact parameters corresponding to scattering angles 60° and 90° respectively, when an a particle is approaching a gold nucleus. For d, =x d,, the value of x will be . [29 June, 2023 (Shift-I)] 12. Choose the correct option from the following options given below: [24 June, 2022 (Shit (a) In the ground state of Rutherford’s model electrons are in stable equilibrium. While in Thomson's model electrons always experience a net — force. (8) Anatom has nearly continuous mass distrib a Rutherford’s model but has a highly non-uniform ‘mass distribution in Thomson's model (c) A classical atom based on Rutherford’s model is doomed to collapse. (d) The positively charged part of the atom possesses most of the mass in Rutherford's model but not in Thomson’s model. X10" m, Take : 13. The ratio for the speed of the electron in the 3° orbit of He’ to the speed of the electron in the 3" orbit of hydrogen atom will be: [25 June, 2022 (Shift-I)] (a) 1:1 (6) 1:2 © @ 21 14, A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of: (Given, Planck's constant = 6.6 10% Js). [27 June, 2022 (Shift-1)] (@) 2.10 * 10Js (b) 1.05 x 10°" Js (2) 3.15 « 10%UIs (d) 4.2 10s 15. Given below are two statements: Statement-I: In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E,) to higher energy orbit (E,) is given as hf= E,~E, Statement-I: The jumping of electron from higher energy orbit (£,) to lower energy orbit £, is associated with frequency of radiation given as f= (E, ~ Eh This condition is Bohr's frequency condition, 16. 17 18. 19, 20, In the light of the above statements, choose te cng answer from the options given below: (27 Sune, 2022 Shiny (@) Both Statement-I and statement-I are true, (6) Both Statement-I and statement-H ate fase, (6) Statement-l is correct but statement-tI is false (A) Statement is incorrect but statementll i tue ‘The momentum of an electron revolving in n* oi, given by: (Symbols have their usual meanings) [25 July, 2022 (Shingy nh nh @) oe OT nh 2nr © on oe In Bohr’s atomic model of hydrogen, let K, Pand Ea the kinetic energy, potential energy and total energy ofte electron respectively. Choose the correct option when ie electron undergoes transitions to a higher level: [24 June, 2022 shinny (a) AIK, Pand E increase, (0) K decreases, P and E increase. (0) P decreases, K and E increase. (A) K increases, P and E decrease. If an electron is moving in the 1 orbit ofthe hydrogen atom, then its velocity (v,) for the 1 orbit is given [17 March, 2021 (Shift-N @ od () ver © vgen @ ye " Imagine that the electron in a hydrogen atom is epic by a muon (1), The mass of muon particle is 207 times that of an electron and charge is equal to the charge ofat electron, The ionization potential of this hydrogen sue will be [18 March, 2021 (Shite (@) 13.6eV (b) 2815.2eV (©) 3312eV (d) 2720V Which level of the single ionized carbon has the $3 energy as the ground state energy of hydrogen atom? [17 Maren, 2021 (Shit () 8 Wi @6 @4 PW SEE PYQs Physies —AA A?considera situation in which reverse biased curent of Mr paticular P ~ N junction increase when itis exposed toa light of wavelength < 621 nm. During this process, gnhancement in carrier concentration takes place due to generation of hole—electton pairs. The value band gap js nearly. [22 July, 2021 (Shift-I1)] @ ey (b) 4ev (9 05eV (@ 26 jna hydrogen atom the electron makes a transition from (n+ 1)® level tothe ri level. If n>> 1, the frequency of radiation emitted is proportional to {2 Sep, 2020 (Shift-IN] 1 @; 1 1 og @ a 33, Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close-to BB Sep, 2020 (Shift-11)] @ 2:1 @ 1:2 (© 10:7 @ 5:7 14. The graph which depicts the results of Rutherford gold foil experiment with a-particles is @: Scattering angle Y¥: Number of scattered a-particles detected (lots are schematic and not to scale) [8 Jan, 2020 (Shift-N)] @ (a) Ty O on oO On © @ Y ¥. : eae oO On O on 4S. The time period of revolution of electron in its ground state orbit ina hydrogen atomis 1.6 10-s. The frequency of evolution of the electron in its first excited state (ins! is 26. Aparticle of mass m moves in circular orbits with potential energy U(r)=Fr, where F is a positive constant and r is its distance from the origin. Its energies are calculated using the Bohr model. Ifthe radius of the particle's orbit is denoted by R and its speed and energy are denoted by vand E, respectively, then for the sf orbit (here His the Planck's constant) [JEE Adv, 2020] (@) Ren" and von? (0) Re and van? 3(e@e ry” i 4m ) mieptye anim (© E= @ el 27. Ahydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 A. The radius of the atom in the excited state, in terms of Bohr radius ay; will (he = 12500 eV—A) [11 Jan, 2019 (Shift-1)] @ 25a, 9a, © 164, @ 4a, 28. A particle of mass m moves in a circular orbit in a central potential field U(r) = kr? . If Bohr’s quantization 2 conditions are applied, radii of possible orbits and energy levels vary with quantum number n as: {12 Jan, 2019 (Shift-1)] (@) evn, Een (b) l ry edln, Ey e— ” (0) menE,on @ ne 29. A He" ions in its first excited state. Its ionization energy is: [9 April, 2019 (Shift-11)] @ 6.04ev © 13.60eV (©) 54.40 eV @ 48.36 eV Hydrogen Spectrum 30. Speed of an electron in Bohr’s 7* orbit for Hydrogen atom is 3.6 x 10° m/s. The corresponding speed of the electron in 3% orbit, in ms is: 18 Jan, 2023 (Shift-I)] (a) (1.8 10%) (©) (15 «10 © G.6x 105, (@ Bax 10 ion of shortest {6 April, 2023 (Shitt-D)] {7 Jan, 2020 (Shif-] | 31+ The energy levels ofan hydrogen atom are shown below: (@ 62105 () 56108 The transition corresponding to emi 78» 10 nem wavelength is ——__ Atoms ST aN32. 34. 35. 36. 31. "FT JEE PYQs Physics aul n=4 Cad. n=3 p—n=2 ——uWl: @ce (@D OB (A A photon is emitted in transition from n=4 to n= 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h= 4 x 10" es): [24 Jan, 2023 (Shift-I1)] (@) 94.1 nm (0) 941 nm (© 974m (@ 993 nm ‘A monochromatic light is incident on a hydrogen sample in ‘ground state. Hydrogen atoms absorba fraction of light and subsequently emit radiation of six different wavelengths. ‘The frequency of incident light is x x 10'’ Hz. The value of xis . Given h=4,25 x 10° eVs) [11 April, 2023 (Shift-1)] Anatom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 600 nm, The net energy absorbed by the atom in this process is n x 10“e¥. The value of m is [Assume the atom to be stationary during the absorption and emission process] (Take h = 6.6 x 10° Js and c=3 x 10" mis ). [13 April, 2023 (Shift-I1)] ‘The ratio of wavelength of spectral lines H7, and H, in the Balmer series is 3. The value of is [8 April, 2023 (Shift-11)] 1f 917 A be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be__A. [10 April, 2023 (Shift-1)] As per given figure A, B and C are the first, second and third excited energy level of hydrogen atom respectively. aod If the ratio of the two wavelengths (: & > 1 } is Gothen the value of n will be {15 April, 2023 (Shift-1)] 38. The wavelength of the radiation emitted is 4, w electron jumps from the second excited state to ther excited state of hydrogen atom. Ifthe electron jumps the third excited state to the second orbit of the hydrop atom, the wavelength ofthe radiation emited wij 20, . The value of is_ 125 Jan, 2023 (Shy x M 39, For hydrogen atom, 2, and 2, are the wavelengy, corresponding to the transitions 1 and 2 respectively shown in the figure. The ratio of 2, and 2, is & yy, u value of xis {51 Jan, 2023 (Shin. 3-————JJ Excited stats m2 —— 2 1 Ground rl state 40. Abeam of monochromatic light is used to excit th electron in Li* from the fist orbit tothe third orbit. The waveengh of monochromatic light is found to be x 10m The value of xis [Given hc = 1242 eV nm] [27 June, 2022 (Shift 441. Find the ratio of energies of photos produced duet transition of an electron of hydrogen atom from is () second permitted energy level to the first level, and (i) the highest permitted energy level to the first permitted level. 29 July, 2022 (Shite) @ (© 1:4 (b) 4:3 @ 4:1 42. Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength 2 The valeot pal quantum number *n’ of the excited state will: pri (R: Rydberg constant) [25 July, 2022 (Shift) AR aR © fa © Ves a aR © Viral © Vig 43. eal is the ratio of energies of photons produced d= itansition of an electron of hydrogen atom from it (i) third permitted energy level to the second level and c > 2 (ii) the highest permitted energy level to the secou! ry permitted level. A The value ofx willbe ___[25 July, 2022 (Sift 44. In the hydrogen spectrum, 2 be the wavelength of first * ransition line of Lyman series, The wavelength difference will be “aA.” between the wavelength of 3" transition line of Paschen series and that of 2" transition line of Balmer series where @ (26 July, 2022 (Shift-1)} 4s. Inthe given, figure, the energy levels of hydrogen atom ‘have been shown along with some transitions marked A, B, CD and E, The transitions 4, B and C respectively represent: {24 Feb, 2021 (Shift-I)] +e ean oy =5 ~0.54eV mtr “aasev n=3 -1.Sle BY Cc 1D n=2 + 1 3.4eV 4 E n=1-L—__Y is.gey (a) The series limit of Lyman series, second member of Balmer series and second member of Paschen series, (8) The first member of the Lyman series, third member of Balmer series and second member of Paschen series. (©) The ionization potential ofhydrogen, second member of Balmer series and third member of Paschen series. (@ The series of Lyman series, third member of Balmer series and second member of Paschen series. 46, According to Bohr atom model, in which ofthe following transitions will the frequency be maximum? [24 Feb, 2021 (Shift-IN)] (@) n=3ton=2 () ton=3 (ce) n=2ton=1 (@) n=Ston=4 4. The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region? [17 March, 2021 (Shift-I1)] (a) Paschen series (0) Balmer series (©) Lyman series (d) Brackett series 48. A particular hydrogen like ion emits radiation of frequency 2.92 x 10! Hz when it makes transition from n=3 ton= 1, ‘The frequency in Hz of radiation emitted in transition from n=2ton= 1 will be: 126 Aug, 2021 (Shift-1)] (@ 246 x 10% (b) 4.38 « 1088 (©) 6.57% 10% (a) 0.44 x 108 49. The wavelength of the photon emitted by a hydrogen atom ‘When an electron makes a transition from n= 2 ton state is: [25 Feb, 2021 (Shift-I1)] (@) 121.8nm (6) 194.8 nm (©) 490.7 om (@ 913.3 nm — 50. Si. 52. 58. 56, 31, IFA, and A, are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of 2, : 4, [26 Feb, 2021 (Shift-1)| (a) 1:3 (6) 7: 108 (1:9 (d) 75135 ‘The recoil speed of'a hydrogen atom after it emits a photon in going from n= 5 state to n= | state will be: [26 Feb, 2021 (Shift-11)| (a) 4.17 m/s (b) 2.19 m/s (©) 434 mis (d) 3.25 ms X different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to states, with principal quantum number n= 6? The value of X is a 127 Aug, 2021 (Shift-I1)] ‘The first three spectral of H-atom in the Balmer series are given 2, 2, A, considering the Bohr atomic model, the ‘ wave lengths of first and third spectral lines (2) are dy related by a factor of approximately ‘x’ *10", [16 March, 2021 (Shift-I)] Which of the following statement(s)is(are) correct about the spectrum of hydrogen atom? |JEE Adv, 2021] (@) The ratio of the longest wavelength to the shortest wavelength in Balmer series is 9/5 (®) There is an overlap between the wavelength ranges of Balmer and Paschen series (© The wavelength of Lyman series are given by (io » Where 2, is the shortest wavelength of Lyman series and mis an integer (@ The wavelength ranges of Lyman and Balmer series do not overlap The energy required to ionise a hydrogen like ion in it ‘ground state is 9 Rydbergs, What is the wavelength of the radiation emitfed when the electron in this ion jumps from, the second excited state to the ground state: {9 Jan, 2020 (Shift-11)} (6) 24.2 nm (@) 8.6 nm ‘The first member of the Balmer series of hydrogen atom hhas a wavelength of 6561 A. The wavelength of the second ‘member of the Balmer series (inn {8 Jan, 2020 (Shift-11) In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is 304A. The corresponding difference for the Paschan series in A is « [4 Sep, 2020 (Shift-1)) @ 4am © 35.80 Atoms 457° ERD seh.58, 59, 60, 61 62. 63, An excited He ion emits two photons in succession, with Wavelengths 108.5 nm and 30.4 nm, in making a transition ‘O ground state, The quantum number n, corresponding to 1s inital excited state is (for photon of wavelength 2, energy £ = 1240e 12 April ‘I — 112 April, 2019 (Shift-1) (@) () n=6 ( n=5 (@ n=7 Inti »¢lectron in first Bohr orbit i excited ta level by a radiation of wavelength 2. When the ion gets deexcited to the ground state in all possible ways(including intermediate emission) a total of six spectral lines are observed. What is the value of 2? (Given: h = 6.63 « 10 jg; = 3% 10% ms) 110 April, 2019 (Shift-11)] (@) 108mm (6) 14m (0) 9.4nm (@) 12.3nm Radiation coming from transition n = 2 to n = 1 of hydrogen atoms fall of He" ions in n = | and n= 2 states, The possible transition of helium ions as they absorb nergy from the radiation is: {8 April, 2019 (Shift-1] (@) n=1on=4 (b) n=25n=4 (c) n=2n=5 (@) n=23n=3 Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 t0 = 2) as 660 nm, the wavelength of the 2nd Balmer line (= 4 to n= 2) will be [9 April, 2019 (Shift-1)] (6) 642.7 nm (4) 388.9 nm The electron ina hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, 2/2. of the photons emitted in this process (a) 889.2 nm (©) 488.9 nm is: [12 April, 2019 (Shift-11)] (a 207 (b) 1s 97 (a 205 Ina hydrogen like atom, when an electron jumps from the M-shell to the L-shell the wavelength of emitted radiation is 2. Ian electron jumps from N-shell to the L-shell the wavelength of emitted radiation will be: [11 Jan, 2019 (Shift-I1) 64, A fice hydrogen atom alter absorbing wavelength 2, gets excited from the state state n= 4, Immediately afer that the elect n= m slate by emitting a photon of wavelen, chang in momentum of atom due to the abso V0 te On jumps 1th, Let he OMioN ad iy t Wh of the option(s) ivare correct? HEE Ady, ayy {Use fe = 1242 eV nm: 1 nm = 10° m, hand ¢ Planck's constant and speed of light, respective) (a) 4,=418 nm (6) The ratio of kinetic enerey ofthe eles iy emission are Ay, and Ay, respectively 19,9, state n= m to the state n= is (c) m=2 1 (@) dpfap,= > 65. Anexcited He ion emits two photons in succession, wig wavelengths 108.5 nm and 30.4 nm, in makings asin 40 ground state, The quantum number, corresponding its initial excited state is (for photon of wavelength, 1240eV = 5 2 April, i eneray Eo Ty) (U2 April 2019 shit (a) n=4 (b) n=6 (0 n=5 (A) n=7 ( Atomic Collision } 66. A free electron of 2.6 eV energy collides with aH ion This results in the formation of a hydrogen atom in thee excited state and a photon is released. Find the fequeny of the emitted photon. (/i = 6.6 x 10 Js) [31 Aug, 2021 (Shift) (a) 0.19 10! MHz (6) 1.45 «10° Miz (ce) 1.45 * 10" MHz (d) 9.0 « 10” MHz 67. A particle of mass 200MeV/e2 collides with a hydrogst ‘lom at rest, Soon after the collision the particle comes!2 ‘est, and the atom recoils and goes to its first excited se 21 16 i ( () 5S ‘The initial kinetic energy of the particle (it His Tw 25, a value of V is: (Given the mass of the hydrogen atom tof © 7G OF | Gee’) [5 Sep, 2020 (Shift JEE PYQs Physics a‘An ideal gas is enclosed in a cylinder at pressure of 2 im and temperature, 300 K. The mean time between two successive collisions is 6 * 10s, Ifthe pressure is doubled and temperature is increased to $00 K,, the mean time between two suc ive collisions will be close to {12 Jan, 2019 (Shift-11y] (0) 4*10%s (@ 3*10%s (a) 2* 107s (0) 05% 108s ¢9. Analpha-particle of mass m suffers 1-deminsional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the mucleus is [12 Jan, 2019 (Shi ID] (a) 2m (b) 3.5m (©) 15m (d) 4m ( X-ray } 0. Thew ‘aves emitted when a metal target is bombarded with high energy electrons are {8 April, 2023 (Shi (0) X-rays (@) Microwaves (©) Infrared rays (d) Radio Waves nm. n The K,,-X ray of molybdenum has wavelength 0.071 nm I the energy of a molybdenum atom with a K electron knocked out is 27.5 ke, the energy of this atom when an 1. electron is knocked out will be __ke V. (Round off to the nearest integer) [27 July, 2021 (Shift-11)] [h= 4.14 «10 eVs,¢=3 * 10° ms] In an X-ray tube, electrons emitted from a filament (cathode) carrying current / hit a target (anode) at a distance d from the cathode. The target is kept at a Potential V higher than the cathode resulting in emission of continuous and characteristic X-rays. If the filament current / is decreased to 4 the potential difference V is increased to 2V, and the separation distance is reduced to S then IJEE Ady, 2020} (a) The cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-rays will remain the same (6) The cutoff wavelength as well as the wavelengths of the characteristic X-rays will remain the same (6) The eut-off wavelength will reduce to half, and the intensities ofall the X-rays will decrease ‘Thecut-off wavelength will become two times larger, and the intensity ofall the X-rays will decrease @ \__ANSWER KEY £136 2.1828) 3 wg @ "Bl 1 Ba 15. 1) 20) 23a) 24, 28, (o) i O 2@ 3 B) 34. (4125) 35, 27) ‘ 2) 43.15} 4 5] as, (d) s. eo 82. [15] 53. [15] 34. (ad) 55, @ “ o SO) Oa) 6K) 64, (be) O) 1 10) 72, (ae) 6 (d) 1 © —8&@ 921 10. jars %@ 126) 1 @ ay, (6) 20, o ' 26 (be) 27.) 28. (a) 25, () 36. [3668] 37. [5] 38, (27) 39, R7] % © 426) a @) gg, @ 56. [486] 7, [10553.33] 58. (c) S66.) 67, 51] 68. (b) oo2 1.(136) Ey =13.62 For Li® Z=3,n=3 (5 E,.=136 xpper Comparing above result with x «10'eV we get x= 136 2. (828]Ground state energy = -13.6 eV eV -13.6412.75 = 085 = fed Angular momentum = ingular mn or 2 24 14210 = Angular momentum x 28710 “a_a_etel 3. (c) Energy used to Bombard gaseous hydrogen, E=12.5eV Energy of excited state = The energy of gaseous hydrogen after bombardment £, =~ 13.6 eV + 125 e¥=-l.be¥ + orbital no of excited state, Vs So, number of spectral lines is nh On Lh on {In second orbit of hydrogen atom (n = 2), lL, ake 2° on So, change momentum = L, ~ L, = 2b 4. (c) Angular momentum, For Bohr orbit is, 1, = 2 “JE PYQs Physics 6 (dy 7. (b) 8. (a) AE, =2.2eV=2.271.6710°%S AE, = 5.2eV = 5.2% 1.67 10S AE, =3eV=3 7 1.6410 AE, = WeV = 107 1.67 10°F 6.6210 «310° 241.6710" = 564.nm =136¢47(1- 4 = AE =40.8eV = 16 Jer We know that, Radius of m* orbit, R & sk11. GB] Weknow that, dx cot 3B. @ 4.) £42510 m a oh =x . (c) According to Rutherford, e—revolves around nucteus in circular orbit. Thus e- is always accelerating (centripetal acceleration). An accelerating chan; emits EM radiation and thus e should loose energy and fall into heavy nucleus collapsing the whole atom, As we know the atomic number He*,Z and the atomic number Hydrogen. H*.Z,, since, voc Z x Z (n= constant) n We have; Vie © Lye (i) and, vy. © Zy ii) Now, On dividing equation (i) and equation (ii) we have; Vy 2, ian Given AE = 10.2 eV For atom in ground state, n, = 1 _AE= nde m 10.2 = 13.6 srease in Angular Momentum, AL = L,~ L 15. (d) 16. (a) 17. () mh 2a th oh x Qn 66x10" 23.4 05x10 “J ~ see When & jump from lower to higher energy level it absorbed energy. Therefore statement | is wrong Forstatement I For the jumping of electron from higher energy orbit (E,) to lower energy orbit (E,) Iw=E,-E, E-& a ah On nh Therefore, momentum, p= mv 2ar Potential energy: An electron posse: es some potential energy because it is found in the field of nucleus potential energy of an electron in n th orbit of radius r, is given by Ke u Where Z = atomic number, = orbit number and dius of the nth orbit Kinetic energy: Electron possesses kinetic energy because of its motion. Closer orbits have greater kinetic energy than outer ones, KZ UU _ ike? 2, Total energy: Total enengy (7) is the sum of potential ‘energy and kinetic energy ie. B= K + U ‘The relationship between PE/TE, and KE is given by: KE =—TE=~PER From above it is clear that as increases, energy inereases, kinetic ‘energy increases, the potential xy decreases and total AtomsRw va! n ob) Batra 26, (he) U © Pr [Using U~ Potential energy and v velocity, nay confusion between theit symbols} Wp > Force = em kom a +5 Magnitude of force » Constant = F (mdeV Uonization potential = 13.6. — um) 13.68 207 eV = 2815.20V comp th 20. (a) Ground state energy of the (H-atom) et an ow Eyam" 13.600" spat Fo” Fit FH an Renew 2 yp 3 ox = -1Mbel n= 6 wn i " nim " 21. (@) Minima required enengy, Ena =A ny nh _( Arm syste aves Famk | 2a ayn NS Ag im 7 2a mt (b) is correct. ex = hv= 28 xls n sve 23. (a) k= qv Lon 27, (©) Energy supplied 2 - = 12500 2. 75ev Rg 980 ave Pe m s E,~EB,= 1215 Yn. [ante 2:1 200 Ve MWe 24, (a) Results of Rutherford gold foil experiment gives ss lisonsd New radius = loa, 28 (a) U Fone, «MU ‘oree, FS tr For circular motion a fu i And mr = : eee VE) JEE PYQs Physics[From quation ()] E=kr Ean 9. (0) Energy levels in Hydrogen like atom is given by. 2a 13.6eV As total energy of He* in 1* excited state i ionization energy should be + 13.6 eV. 136 eV, =2x3.6x108 m/s 3 = 84x10 m/s. he AE pax For shortest wavelength, energy gap should be maximum. Me) Don So, correct choice is transition from n= 3 10.7 =13.6¢ 94.1 nm 33. [3] Let the number of specteal time be n=l) Wan 12=0 dnt 3n-12=0 nin—4) +3 (04) n=4.n=—3 (ignore) hw ave self H 4.25x10" =3< 10 Hz MM. [4125) = hel +) } (4125) = i =66x10™ 3.104 600% 10 500x107 = 4125x107 = 4125 x 10401" 35, [27] For hydrogen atom = For balmer seri And for Hm, (:-3] aH eed iol = a 36 3R _ ft 1) ar 16 stanlt-t | -3k 16 ay E 16] 16 7-39 4s 36, dy SR a x i, 20 x=27 36. [3668] For Lyman series 1 x = re Lowest For Balmer series 3668 A5 as (i) (33) : For B to C,n,=3,n,=4 1 I RO") S-aa|=R [je ay Dividing (ii) by (i) A 7/144 7 7 %, 5/3620. 4x5. n=5 38, [27] —————_n=4 a-3 n=2 n=l Electron jumps from second excited state —> first excited state i 36-7) Ai) For the electron to move from third excited state > second orbit n=4-9n=2 he 1 —— = 13.6] > = 00, /x) (= Dividing equ (ii) by (i) 39.27] Ai) ool) Dividing eqn (ii) by (i) dy 27 a2 ax=27 40, [114]Z= tee x 42. (0) According to definition of wave number, wecan ye fh : a —t— n=l ED JE PYQs PhysicsFor 3* line of Paschen series. n, =3,,=6 Gl 3k 48. (@) A: For transition 4, n, Lyman series 0 => Series limit of B: For transition B, n, = 2, n= 5 => Third line of Balmer series C: For transition C, n, = 3, n, Paschen series = Second line of 46. (©) We know that, AE Ifm,=3 and n, (+4 9-4 5 FF) 36 36 lfm =4andn,=3 (4 -16_ 9 400 400 Since, AE = hv. Hence option (c) is the correct answer. 47. (b) Balmer series lies in the visible region. +3. =109677.6em"* 13] x 1 =121 9593" = '2160m 50. (d) The wavelength of spectral line of the third member of Lyman series; f z(t F a The wavelength of spectral line of the first member ‘of paschen series; RZ (¢ +) ry SI. (a) AS ° AE = 13.60.54 AE= 13.06 eV a0) Using conservation of linear momentum, Atoms«when 1, = 3 he ee = AE 13.06%1.6%10" yan om, X10GTKIOT 52. [15] No. of different wavelengths _Mx(n=N) _ 6x5 _ 2 2 : 364.5nm_ Hence, for the Balmer series, 36/5R _9 a 4/R 5 For the Paschen series, n, =3 and n,= 4, 53.(15] length of first and third spectral lines 22136 100 5 512= 15x10" 144 => egy =e = 1874.7, mx aR in» WHEN My Fosigy <2 =82020m where R = 1.0973 « 107 m' = Rydberg constant. 5 and n, = 2,3,4, 55. (a) ee (13.6e' -as6evys') he 7 e ar =(13r)(3) Wavelength = 42 nm = 11.39 nm 8x13.6 n integer not an integer For the Balmer series, n, = 2 and n, = 3, 4, 5, 6, “1 JEE PYQs Physics20 rr 2 — «6561 A = 481 7 160A, Hence, wavelength of the second member of the Balmer series is 486 nm. 1 {10553.33] From, Bohr’s atom model for hydrogen atom, oo) 81x3 304 =10553,14A 1240 1240 1085 304 1240eV _15 ——— == «9% 13.60 e316 10.8.nm 61. 60. (b) Energy released for transition (n = 2 ton =!) =13.6x2xeV 4 For He’ ions Z=2 Using option (@n=lton=4 exo. ()n=2ton=4 E=13.6x () n=2ton=5 E=13.6x2 @n=2ton=3 Equation (i) = (ii) 2 _SxI6_ _ 660%5x16 G60 3603 PO seg E88 IH5 Ay 365x916 _ 20 Meee e967 7 916 63. (d) For ML steel Ail) =>m=2 From (ii) M136 tte 13.6.2ev he ? 16 124216 nm We have KE, w “D> JEE PYQs Physics 65. (c) 1240, 1240 108.5 30.4 54.4 PE. = 0 [For every large distance] «Total energy = 2.6 + 0= 2.6 eV 66. (b) In Ist excited state of H, total energy =-3.4ey ‘Thus loss in total energy = 6 eV Energy of Frequency of emitted photon = ae 6x1.6x10" 3 = SO = 145 x10 Hz = 145x107 66x10 67.[51] M,= 200MeVIC*, m= 1GeviC* x, Final velocity of hydrogen atom is “7 MV,= mV Initial velocity of particle Mol m v Also, Latgy? =Lmv? + 313.6 2 4 Put (i) and get answer 1 31 Mo we =a Hence, N= 51. 68. (b) 2, 500 2p Y300 lo* = x =anotse4g, (d) bet the unknown mass is Mt. 71. (10] Given, h=4.14 10" eVs,c=3 * 10" m/s Oo ©)-O © ™ (Before) (Aner) 1242x107 0071x107 a. Energy of the atom E, = 27.5- (i) my, =-m, + Mo, Gi y= +9, E,=245~17.5=10 keV After solving above equation, M= 4 m 70. (b) When a target metal is bombarded with high energy electron the X rays are emitted. Hence / decreases