BOHR S THEORY AND PHYSICS OF ATOM CHAPTER 43

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1 1. a BOHR S THEORY AND PHYSICS OF ATOM CHAPTER 3 1 h A T (ML T ) M L T 3 L me L MLT M(AT) M L T a has dimensions of length.. We know, 1/ (1/n 1 1/n ) a) n 1, n 3 or, 1/ (1/ 1/9) 36 or, nm b) n 1, n 5 1/ (1/16 1/5) or,. 1 m. nm for R 1.9 1, 5 nm c) n 1 9, n 1 1/ (1/81 1/1) 81 or, nm for R ; nm 3. Small wave length is emitted i.e. longest energy n 1 1, n 1 1 a) R n 1 n b) z R n 1 n 1 91 nm 3 nm 1.11 z 1 1 c) z R n 1 n 91 nm 91 1 nm z 9. Rydberg s constant me 3 8h C nm. m e kg, e c, h J-S, C m/s, or, R (1.6 1 ) ( ) 3 1 ( ) 5. n 1, n E 13.6 n1 n n1 n 13.6 (1/ 1/) 13.6/ 3. ev m 1

2 6. a) n 1, r h n mze A.53n A Z 13.6z ev n b) n, r. A ev c) n 1, r 6.5 A A 1. As the light emitted lies in ultraviolet range the line lies in hyman series R n 1 n (1/1 1/n ) (1 1/ n ) (1 1/ n ) n n n a) First excitation potential of He + 1. z 1..8 V ++ b) Ionization potential of L V z V 9. n 1 n n nm n 1 and n nm n 1 3 n

3 nm A E hc The transition takes place form n 1 to n Now, ex / z z z The ion may be Helium. q1q 11. F r [Smallest dist. Between the electron and nucleus in the radius of first Bohrs orbit] (1.6 1 ) (1.6 1 ) (.53 1 ) N a) From the energy data we see that the H atom transists from binding energy of.85 ev to exitation energy of 1. ev Binding Energy of 3. ev..85 ev So, n to n b) We know 1/ ev (1/ 1/16) 3. ev nm ev The second wavelength is from Balmer to hyman i.e. from n to n 1 n 1 to n R n 1 n nm Energy at n 6, E.3 36 Energy in groundstate 13.6 ev Energy emitted in Second transition 13.6 ( ) ev b) Energy in the intermediate state 1.13 ev z n n 13.6 or, n n The potential energy of a hydrogen atom is zero in ground state. An electron is board to the nucleus with energy 13.6 ev., Show we have to give energy of 13.6 ev. To cancel that energy. Then additional 1. ev. is required to attain first excited state. Total energy of an atom in the first excited state is 13.6 ev ev. 3.8 ev.

4 Energy in ground state is the energy acquired in the transition of nd excited state to ground state. As nd excited state is taken as zero level hc E ev Again energy in the first excited state E II 15 8 hc ev. 1. a) The gas emits 6 wavelengths, let it be in nth excited state. n(n 1) 6 n The gas is in th excited state. n(n 1) b) Total no.of wavelengths in the transition is 6. We have 6 n. 18. a) We know, m r mr w w hn m r (.53) rad/s rad/s. 19. The range of Balmer series is nm to 365 nm. It can resolve and + if / No.of wavelengths in the range 36 8 Total no.of lines [extra two is for first and last wavelength]. a) n 1 1, n 3, E 13.6 (1/1 1/9) /9 hc/ or, nm b) As n changes by, we may consider n to n then E 13.6 (1/ 1/16).55 ev and.55 1 or 8 nm. 1. Frequency of the revolution in the ground state is V r [r radius of ground state, V velocity in the ground state] V Frequency of radiation emitted is r f C f C/f C r V Cr V nm 5. nm.. KE 3/ KT 1.5 KT, K ev/k, Binding Energy 13.6 (1/ 1/1) 13.6 ev. According to the question, 1.5 KT T T K No, because the molecule exists an H + which is impossible. 3. K ev/k K.E. of H molecules 3/ KT Energy released, when atom goes from ground state to no (1/1 1/9) 3/ KT 13.6(1/1 1/9) 3/ T T K.

5 . n, T 1 8 s Frequency me 3 3 n h So, time period 1/f 3 3 o n h me sec. 8 1 No.of revolutions revolution. 5. Dipole moment () n i A 1 q/t A qfa (8.85) (6.63) (1.6) me me ( r n ) e ( r n ) h n h n (9.11 )(1.6 1 ) (.53) ( ) (6.6 1 ) (1) A - m. 6. Magnetic Dipole moment n i A n 3 3 e me r n h n 3.5 Angular momentum mvr Since the ratio of magnetic dipole moment and angular momentum is independent of Z. Hence it is an universal constant h n e m r n (1.6 1 ) (9.11 ) (3.1) (.53 1 ) Ratio 1 3 ( ) ( ) C/kg.. The energies associated with 5 nm radiation ev Energy associated with 55 nm radiation ev. 55 The light comes under visible range Thus, n 1, n 3,, 5, E E (1/ 1/3 ) 1.9 ev E E 13.6 (1/ 1/16).55 ev E E (1/ 1/5).856 ev Only E E comes in the range of energy provided. So the wavelength corresponding to that energy will be absorbed nm 8 nm.55 8 nm wavelength will be absorbed. 8. From transitions n to n 1. E 13.6 (1/1 1/) / 1. ev Let in check the transitions possible on He. n 1 to E (1 1/).8 ev [E 1 > E hence it is not possible] n 1 to n 3 E 13.6 (1 1/9) 8.3 ev [E > E hence impossible] Similarly n 1 to n is also not possible. n to n 3 E (1/ 1/9).56 ev

6 n to n E 13.6 (1/ 1/16) 1. ev As, E 3 < E and E E Hence E 3 and E can be possible nm Work function Energy required to remove the electron from n 1 1 to n. E 13.6 (1/1 1/) 13.6 hc 13.6 KE KE KE ev nm hc 1 E 1. ev 1 a) The possible transitions may be E 1 to E E 1 to E, energy absorbed 1. ev Energy left ev hc 1. ev or nm E 1 to E 3, Energy absorbed 1.1 ev Energy left ev hc 1.3 or nm.3 E 3 to E, Energy absorbed.65 Energy left ev hc or b) The energy absorbed by the H atom is now radiated perpendicular to the incident beam. 1. hc 1.1 hc or 1 1. or nm 1.6 nm.65 hc 1 or nm ev a) The hydrogen is ionized n 1 1, n Energy required for ionization 13.6 (1/n 1 1/n ) 13.6 hc nm 8 nm. b) For the electron to be excited from n 1 1 to n E 13.6 (1/n 1 1/n ) 13.6(1 ¼) hc / nm. 3. The given wavelength in Balmer series. The first line, which requires minimum energy is from n 1 3 to n. The energy should be equal to the energy required for transition from ground state to n 3. i.e. E 13.6 [1 (1/9)] 1.9 ev Minimum value of electric field 1.9 v/m 1.1 v/m

7 33. In one dimensional elastic collision of two bodies of equal masses. The initial velocities of bodies are interchanged after collision. Velocity of the neutron after collision is zero. Hence, it has zero energy. 3. The hydrogen atoms after collision move with speeds v 1 and v. mv mv 1 + mv (1) mv mv1 mv E () 1 1 From (1) v (v 1 + v ) v v v v 1 From () v v v E / m v1v E m (v v ) (v v ) v v (v 1 v ) v E/m For minimum value of v v 1 v v (E/m) v v E m (3). 1 m/s. 35. Energy of the neutron is ½ mv. The condition for inelastic collision is ½ mv > E E ¼ mv E is the energy absorbed. Energy required for first excited state is 1. ev. E < 1. ev 1. ev < ¼ mv V min m v 6 1 m/sec a) nm ev Momentum P E/C hc 1 h 3 c kg-m/s b) v v 1/ m/s c) KE of atom ½ (.6) Difference in energy in the transition from n 3 to n is 1.89 ev. Let recoil energy be E. ½ m e [V V 3 ] + E 1.89 ev J ev ev E J 3 E n 1, n 3 Energy possessed by H light 13.6 (1/n 1 1/n ) 13.6 (1/ 1/9) 1.89 ev. For H light to be able to emit photoelectrons from a metal the work function must be greater than or equal to 1.89 ev.

8 39. The maximum energy liberated by the Balmer Series is n 1, n E 13.6(1/n 1 1/n ) / 3. ev 3. ev is the maximum work function of the metal.. Wocs 1.9 ev The radiations coming from the hydrogen discharge tube consist of photons of energy 13.6 ev. Maximum KE of photoelectrons emitted Energy of Photons Work function of metal ev 1.9 ev 11. ev 1. nm, e Charge of an electron, ev, V stopping potential hc We have, ev 9 1 ev.83 ev V.83 volts.. Mass of Earth Me 6. 1 kg Mass of Sun Ms. 1 3 kg Earth Sun dist m mvr or, m v r n h (1) GMeMs Mev or v GMs/r () r r Dividing (1) and () ev ev + We get Me r for n 1 n h GMs r h GMsMe m m. b) n Me r G Ms.5 1. h 3. m e Vr z (1) GMnMe mev GM n () r r r Squaring () and dividing it with (1) e n h r Gmn m v r KE PE rm e GMnMe e M n h me r from (1) GMnM n h r Gm e n r 3 (GMnM e ) G MnMe e me 1 1 m V n h 3.8 n h r n Gm me 3 n e n e n e n e GM M GM M GM M G M M r n h n h 3 e G MnM Total energy KE + PE n h

9 . According to Bohr s quantization rule mvr r is less when n has least value i.e. 1 or, mv (1) R Again, r mv, or, mv rqb () qb From (1) and () rqb [q e] r r eb r h / eb [here n 1] b) For the radius of nth orbit, r c) mvr, r mv qb Substituting the value of r in (1) mv mv qb eb m v [n 1, q e]. eb heb heb v or v. m m 5. even quantum numbers are allowed n 1, n For minimum energy or for longest possible wavelength E n1 n.55 hc hc nm 8 nm Velocity of hydrogen atom in state n u Also the velocity of photon u But u << C Here the photon is emitted as a wave. So its velocity is same as that of hydrogen atom i.e. u. According to Doppler s effect 1 u / c frequency v v 1 u / c as u <<< C u 1 q c 1 u / c u v v v 1 1 c v u v 1 c 3.9

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