Tiêu đề 12. ATOMS.pdf ✅

6. () : Explana on According to Bohr’s postulate The de-Broglie wavelength 7. () : Explana on The energy of photons is given by Where, is Rydberg constant, is Planck’s constant and is the speed of light Energy of photon produced from second to first energy level Energy of photon produced from higher en‐ ergy level (i.e., ) to second level 8. () : Explana on is for con nuous -rays (where is voltage) The values in characteris c - rays spec‐ trum (such as . ) do not change with . It's also essen al to understand that only a por on of the kine c energy is trans‐ formed into -ray energy. So correct op on is (2). 9. () : Explana on 10. () : Explana on Conceptual Ques on 11. () : Explana on When an electron of the beam, a er hi ng a hydrogen atom, completely looses its energy, the hydrogen atom jumps to the maximum excited state and its energy becomes . . So the hydrogen atom jumps to the second excited state. Total number of spectral lines . 12. () : Explana on We know that, So, for (1) Similarly, for (2) Hence, from Eqs. (1) and (2), we get 13. () : Explana on Op on (1) is correct. 14. () : Explana on From the formula, 15. () : Explana on We know wavelength of emi ed photon is given by , where is the energy difference between two stable energy levels. From the given energy levels we can say that transi on , corresponds to emission of shortest wavelength because shortest wave‐ length corresponds to highest energy differ‐ ence (de-excita on) which will occur in , as to . So correct op on is (1). 2πr = nλ ∴ λ = = 2πr 2πr n (∵ n = 1 at ground state) ⇒ λ = 2 × 3.14 × 0.53 0 A = 3.33 0 A E = Rhc ( − ) 1 n 2 1 1 n 2 2 R h c E1 = Rhc ( − ) 1 1 2 1 2 2 E1 = Rhc 3 4 ∞ E2 = Rhc ( − ) = Rhc 1 2 2 1 ∞ 1 4 ∴ = = or3 : 1 E1 E2 Rhc 3 4 Rhc 1 4 3 1 λcutoff = hc eV0 X V0 λ X λkα, λkβ,... V0 X E = = (Ze)(2e) 4πε0r ( 9 × 10 9 × 79× 2(1.6 × 10 −19) 2 ) 41.3 × 10 −15 = J 9×79×3.2×1.6×10−14 41.3 = 8.814 × 10 −13Joule = 8.814×10 −13 1.6×10 −13 E = 5.51 MeV . (−13 ⋅ 6 + 12.1)ev i. e. −1 ⋅ 5 ev ∴ − = −1 ⋅ 5 ⇒ n = 3 13 ⋅ 6 n2 ∴ N = = = 3 n(n − 1) 2 3(3 − 1) 2 = R [ − ] 1 λ 1 n 2 1 1 n 2 2 λ1 ⇒ = R [ − ] 1 λ1 1 (3) 2 1 (4) 2 = R [ ] 1 λ1 7 144 λ2 ⇒ = R [ − ] 1 λ2 1 (2) 2 1 (3) 2 = R [ ] 1 λ2 5 36 = λ1 λ2 20 7 eV = ⇒ V = hc λ hc eλ = 6.6×10 −34×3×10 8 1.6×10 −19×0.4125×10 −10 = 30 × 10 3 V = 30kV λ = hc ΔE ΔE D D n = 3 n = 1
16. () : Explana on As we know, Therefore, (1) (2) from eq.(1) and eq.(2), we get 17. () : Explana on The electron moves around the nucleus due to centripetal force provided by the electro‐ sta c force, which is given by Also, radius of orbit, From eq.(1) and eq.(2), we get 18. () : Explana on A set of atoms in an excited state decays in general to any of the states with lower energy. 19. () : Explana on First ioniza on energy of helium atom is Second ioniza on energy of helium atom is Energy required to remove both the elec‐ trons from helium atom is 20. () : Explana on For situa on at sub atomic level also basic laws are applicable. 21. () : Explana on Wavelength in Balmer series of hydrogen spectrum is given by rela on where So, wavelength of first line Similarly wavelength of second line, From eq,(1) and eq,(2), we get Hence, the wavelength of the second line in the same series is 22. () : Explana on The maximum number of photon emissions Photons are emi ed when electron have a transi on from higher to lower state. The minimum is 1 for a direct jump. So correct op‐ on is (3). 23. () : Explana on For the distance of closest approach, we can write, Where is the distance of closest approach. 24. () : Explana on Posi ve nucleus exhibits a rac ve coulomb force on electrons. Bohr's atomic model pro‐ posed electrons orbi ng the nucleus without emi ng radia on as long as they're in fixed (sta onary) orbits. So correct op on is (3). v = nλ ⇒ = ⇒ = R ( − ) 1 λ n v 1 λ 1 n 2 1 1 n 2 2 ⇒ v = Rc ( − ) 1 n 2 1 1 n 2 2 v2 = Rc ( − ) = Rc ( − ) 1 2 2 1 3 2 1 4 1 9 v1 = Rc ( ) = 1 2 2 Rc 4 v3 = Rc ( ) = 1 3 2 Rc 9 ⇒ v1 − v3 = Rc ( − ) 1 4 1 9 v1 − v3 = v2 ⇒ v1 − v2 = v3 Fe = ⇒ Fe ∝ (1) Ze 2 4πε0r 2 n 1 r 2 n rn = ( ) h 2ε0 πme 2 n 2 Z ⇒ rn ∝ n 2 (2) Fe ∝ ⇒ Fe ∝ n 1 −4 (n2) 2 I1 = 24.6 eV I2 = = 54.4eV 13.6(2) 2 (1) 2 ∴ = I1 + I2 = 24.6 eV + 54.4 eV = 79 eV = R [ − ] 1 λ 1 2 2 1 n2 n = 3, 4, 5 = R [ − ] 1 λ1 1 4 1 3 2 = 5R 36 ⇒ R = (∵ λ1 = λ, given) (1) 36 5λ = R [ − ] = 1 λ2 1 4 1 4 2 3R 16 ⇒ λ2 = (2) 16 3R λ2 = λ 20 27 λ 20 27 Nmax = n(n − 1) 2 = = 6 4(4 − 1) 2 mv 2 = ⇒ rα 1 2 K×(2e)(Ze) r 1 m r