Tiêu đề 12.ATOMS - Explanations.pdf ✅

1 (c) For Balmer series, n1 − 2, n2 = 3 for 1st line and n2 = 4 for second line λ1 λ2 = ( 1 2 2 − 1 4 2 1 2 2 − 1 3 2 ) = 3/16 5/16 = 3 16 × 36 5 = 27 20 λ2 = 20 27 λ1 = 20 27 × 6561 = 4860 Å 2 (b) Energy released = E4 − E1 = − 13.6 4 2 − (− 13.6 1 2 ) = 1.75eV 3 (d) E = E4 − E3 = − 13.6 4 2 —(− 13.6 3 2 ) = −0.85 + 1.51 = 0.66 eV 4 (b) For Balmer series, nf =2 and ni=3,4,5,.... Frequency, of 1st spectral line of Balmer series f = RZ 2 c ( 1 2 2 − 1 3 2 ) or f = RZ 2 c × 5 36 ...(i) Frequency, of 2nd spectral line of Balmer series f’ = RZ 2 c ( 1 2 2 − 1 4 2 ) or f′ = RZ 2 c × 3 16 ....(ii) Form eqs. (i) and (ii), we have f f′ = 20 27 ∴ f ′ = 27 20 f = 1.35 f 5 (a) Incandescent electric lamp produces continuous emission spectrum whereas mercury and sodium vapour give line emission spectrum. Polyatomic substances such as H2, CO2 and KMnO4 produces band absorption spectrum. 7 (a) Electron angular momentum about the nucleus is an integer multiple of h 2π , where h is Planck’s constant. Iω = mvr = nh 2π r ∝ n 8 (b) When electric discharge is passed through mercury vapour lamp, eight to ten lines from red to violet are seen in its spectrum. In some line spectra there are only a few lines, while in many of them there are hundreds of them. Hence, mercury vapour lamp gives line spectra. 10 (a) E = E2 − E1 = − 13.6 2 2 — (− 13.6 1 2 ) = 10.2 eV 11 (a) Radius of orbit rn = n 2h 2 4 π2 k 2me 2 rn ∝ n 2 Energy E = −Rch Z 2 n2 E ∝ 1 n 2 12 (b) Least energy of photon of Balmer series is obtained when an electron jumps to 2nd orbit from 3rd orbit. E = E3 − E2 = [ −13.6 3 2 − ( −13.6 2 2 )] eV = 13.6 [ 1 4 − 1 9 ] = 13.6 × 5 36 eV = 1.89 eV 13 (c) In Balmer series, n = 2 E = 13.6 2 2 = 3.4 eV 14 (c) For an electron to remain orbiting around the nucleous, the angular momentum (L) should be an integral multiple of h/2π. ie, mvr= nh 2π where n = principle quantum number of electron, and h= Planck’s constant 15 (c) Energy of helium ions. En = − 13.6 Z 2 n2 eV In minimum position, n=1 For He +, Z = 2 E = −13.6 × (2) 2 1 eV E = 54.4 eV 16 (c) These photons will be emitted when electron makes transitions in the shown way. So, these transitions is possible from two or three atoms. From three atoms separately. 17 (c) As U = 2E,K = −E Also, E = − 13.6 n2 eV Hence, K and U change as four fold each.
19 (c) λ ∝ n 2 ∴ λLyman λBalmer = ( 1 2 ) 2 = 1 4 =0.25 20 (b) Number of spectral lines = n(n−1) 2 = 3(3−1) 2 = 3 21 (d) In Lyman series, wavelength emitted is given by 1 λ = R [ 1 1 2 − 1 n2 ] where, n =2,3,4...... and R =Rydberg’s constant = 1.097 × 107m−1 For maximum wavelength n=2 ∴ 1 λmax = 1.097 × 107 [ 1 1 2 − 1 2 2 ] 1 λmax = 1.097 × 107 [ 1 1 − 1 4 ] = 1.097 × 107 × 3 4 ⇒ λmax = 4 3.291 ×107 = 1216 Å = 121.6 m ∴ λmax =122nm 22 (d) PE= 2 × total energy = 2(−1.5) eV = −3.0 eV 23 (d) Number of spectral lines obtained due to transition of electrons from nth orbit to lower orbit is, N = n(n−1) 2 I case 6 = n1(n1−1) 2 ⇒ n1 = 4 II case 3 = n2(n2−1) 2 ⇒ n2 = 3 Velocity of electron in hydrogen atom in nth orbit vn ∝ 1 n vn v ′ n = n2 n1 ⇒ n6 n3 = 3 4 24 (d) v̅= R [ 1 2 2 − 1 4 2 ] = 3R 4 = 20397cm−1 For the same transaction in He atom (Z = 2) v̅= RZ 2 [ 1 2 2 − 1 4 2 ] = 3R × 2 2 4 = 20397 × 4 = 81588 cm−1 25 (c) The Spectrum of light emitted by a luminous source is called the emission Spectrum. Neon bulb gives an emission Spectrum. The spectrum of the neon light has several bright lines. The red lines are bright. The emission Spectrum of an element is the exact opposite of its absorption Spectrum, that is, the frequencies emitted by a material when heated are the only frequencies that will be absorbed when it is lighted with a white light. Hence, neon sign does not produce an absorption Spectrum. 26 (d) The magnetic moment of the ground state of an atom is μ = √n(n + 2)μB Where, μB is gyromagnetic moment. Here, open sub-shell is half-filled with 5 electrons. ie, n=5 ∴ μ = √5(5 + 2). μB =μB√35 27 (a) 1 λmin = R [ 1 2 2 − 1 3 2 ] = R × 5 36 1 λmax = R [ 1 2 2 − 1 ∞ ] = R 4 λmin λmax = R × 5 36 × 4 R = 5 9 28 (b) ∆λ = 706 − 656 = 50 nm = 50 × 10−9m, v =? As ∆λ λ = v c ∴ v = ∆λ λ × c = 50 × 10−9 656 × 10−9 × 3 × 108 = 2.2 × 107ms −1 29 (c) From mvr = nh 2π v = nh 2πmr Acceleration, a = v 2 r = n 2h 2 4π2m2r 3 = h 2 4π2m2r 3 (n = 1) 30 (d) In the first case, energy emitted, E1 = 2E − E = E In the second case, energy emitted E2 = 4E 3 − E = E 3
As E3 is 1 3 rd, λ2must be 3 times, ie, 3λ 31 (a) Linear momentum = mv = mcZ 137 n Angular momentum = nh 2 π Given, Linear momentum × angular momentum ∝ n x ∴ mcZ 137 n × nh 2π ∝ n x n 0 ∝ n x ⇒ x = 0 32 (b) Given, v = 2.18 × 106 ms−1 , r = 0.528 × 10−10m Acceleration of electron moving round the nucleus a = (2.18 ×106) 2 0.528 ×10−10 ≈ 9 × 1022 ms −2 33 (a) λB λL = ( 1 1 2 − 1 2 2 ) ( 1 2 2 − 1 3 2 ) = 3/4 5/36 = 27 5 λL = 5 27 λB = 5 27 × 6563 = 1215.4 Å 34 (b) The wavelength of series for n is given by 1 λ = R ( 1 2 2 − 1 n2 ) were R is Rydberg’s constant. For Balmer series n=3 gives the first member of series and n=4 gives the second member of series. Hence, 1 λ = R ( 1 2 2 − 1 3 2 ) 1 λ1 = R ( 5 36) ...(i) 1 λ2 = R ( 1 2 2 − 1 4 2 ) = R ( 12 16 × 4 ) = 3R 16 ...(ii) ⇒ λ2 λ1 = 16 3 × 5 36 = 20 27 λ2 = 20 27 λ (∵ λ1 = λ) 35 (c) Given, E2 − E1 = 2.3 eV Or v= E2−E1 h = 2.3 × 1.6×10−19 6.6 ×10−34 = 0.55 × 1015 = 5.5 × 1014 Hz 37 (a) When a γ − ray photon is emitted then atomic number and mass number remains unchanged. 38 (c) The given type of spectrum has coloured bands of light on a dark-ground. One end of each band is sharp and bright and the brightness gradually decreases towards the other end. Band spectrum is obtained from the molecules in the gaseous state of matter. For example, when discharge is passed through oxygen, nitrogen or carbon dioxide, the light emitted from these gases give band spectrum. 39 (d) The moment of linear momentum is angular momentum L = mvr = nh 2π Here, n=2 ∴ L = 2h 2π = h π 41 (a) E = − Z 2 13.6 n 2 eV For first excited state, E2 = −3 2 × 13.6 4 = −30.6 eV Ionisation energy for first excited state of Li 2+ is 30.6 eV. 42 (b) The radius of the orbit of the electron in the nth excited state re = n 24πε0h 2 4π2mZe2 For the first excited state n = 2 , Z = 1 ∵ r’ = 4ε0h 2 πme2 For the ground state of hydrogen atom n = 1 , Z = 1 ∵ r’’ = h 2ε0 πme2 The ratio of radius r ′ r ′′ = 4 1 The ratio of area of the electron orbit for hydrogen atom A ′ A′′ = 4π(r′) 2 4π(r′′) 2 A ′ A′′ = 16 1 43 (c) The excitation energy in the first excited state is E = RhcZ2 ( 1 1 2 − 1 2 2 ) = (13.6 eV) × Z 2 × 3 4 ∴ 40.8 = 13.6 × Z 2 × 3 4 ⇒ Z = 2 So, the ion in problem is He +. The energy of the ion in the ground state is E = RhcZ2 1 2 = 13.6 × 4 = 54.4 eV Hence, 54.4 eV is required to remove the electron from the ion.
44 (b) r ∝ n 2 rf ri = ( nf ni ) 2 21.2 ×10−11 5.3×10−11 = ( n 1 ) 2 n 2 = 4 n = 2 45 (d) Given , En = 13.6 n2 eV Energy of photon ejected when electron jumps from n=3 state to n=2 state is given by ∆E = E3 − E2 ∴ E3 = − 13.6 (3) 2 eV = − 13.6 9 eV E2 = − 13.6 (2) 2 eV = − 13.6 4 eV So, ∆E = E3 − E2 = − 13.6 9 − (− 13.6 4 ) = 1.9 eV (approximately) 47 (d) Kinetic energy of electron K = Ze2 8πε0r Potential energy of electron U = 1 4πε0r Ze2 r ∴ Total energy E = K + U = Ze2 8πε0r − Ze2 4πε0r Or E = Ze2 8πε0r Or E = −K Or K = −E = −(−3.4) Or = 3.4 eV 48 (d) Nucleus Contains only the neutrons and protons. 49 (c) According to the Bohr’s theory the wavelength of radiations emitted from hydrogen atom given by 1 λ = R [ 1 n1 2 − 1 n2 2 ] ⇒ λ = n1 2n2 2 (n2 2n1 2)R For maximum wavelength ifn1 = n,then n2 = n + 1 ∴ λ is maximumforn2 = 3 and n1 = 2. 50 (a) According to kinetic interpretation of temperature Ek = (= 1 2 mv2) = 3 2 kT Given : Ei= 10.2 eV = 10.2 × 1.6 × 10−19 J So, 3 2 kT = 10.2 × 1.6 × 10−19 J Or T = 2 3 × 10.2 × 1.6 ×10−19 k = 2 3 × 10.2 × 1.6 ×10−19 1.38 ×10−23 = 7.9 × 104 K 51 (b) The minimum energy needed to ionise an atom is called ionisation energy. The potential difference through which an electron should be accelerated to acquire this much energy is called ionisation potential. (E2 )H – (E1 )H = 10.2 eV or (E1 )H 4 − (E1 )H = 10.2 eV ∴ (E1 )H = − 13.6 eV Hence ,ionisation potential energy is = (E∞)H − (E1 )H = 13.6 eV ∴ Ionisation potential = 13.6 V 52 (d) Lowest orbit is n = 1. Three lower orbits correspond to n = 1. 2. 3 ∴ E1 = 13.6 1 2 = 13.6 eV , E2 = 13.6 2 2 = 3.4 eV, E3 = 13.6 3 2 = 1.5 eV 53 (a) For Lyman series, n1 = 1, n2 = ∞ 1 λ = R ( 1 n1 2 − 1 n2 2 ) = R ( 1 1 2 − 1 ∞ ) = R 54 (a) For nth Bohr orbit, r = ε0n 2h 2 πmZe 2 de-Broglie wavelength λ = h mv Ratio of both r and λ, we have r λ = ε0n 2h 2 πmZe 2 × mv h = ε0n 2hv πZe 2 But v = Ze2 2hε0n for nth orbit Hence, r λ = n 2π 55 (d) Infrared radiation corresponds to least value of ( 1 n1 2 − 1 n2 2) ,ie, from Paschen, Brackett and Pfund series. Thus the transition corresponds to 5 → 3. 56 (d) As is known, PE = −2KE ie, EP = −2EK or Ep Ek = −2 57 (a) U = eV = eV0ln ( r r0 ) ∴ |F| = |− dU dr | = eV0 r