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Page 1 ¢ Objective Questions (Level 2) 1. F F a c = Þ GmM r mv r 2 2 = …(i) and mvr nh = 2p …(ii) From Eq. (ii) Þ v nh mr = 2p Putting this value in Eq. (i) GM r h h m r = 2 2 2 2 2 4p Þ r n h m GM = 2 2 2 2 4 p KE = = ´ = 1 2 1 2 2 2 mv m GM r GMm r PE =  GMm r Þ E = + KE PE Þ E GMm r GMm m GM n h =  =  ´ 2 4 2 2 2 2 2 p Þ E G M m n h = 2 2 2 2 3 2 2 p for ground state n = 1 Þ E G M m n =  2 2 2 2 3 2 p Hence correct option is (b). 2. We have m p n n n n n i A e T r = = ´ 2 Þ m p p n n n n n n e r u r eu r = ´ ´ ´ = 2 2 2 Þ m n e v n r n ev r n = ´ ´ ´ = ´ 1 1 2 1 1 2 2 Þ m 2 1 1 2 2 = ´ ev r and m 1 1 1 2 1 = ´ ev r Þ m m 1 2 2 = Hence magnetic moment decreases two times correct option is (b). 3. For Hlike atom E n n =  (13.6)Z eV 2 2 Here E z 2 2 4 = (13.6) and E z 1 2 =  13.6 E E 2 1  = 40.8 eV Þ 13.6 40.8 Z 2 1 1 4 ´  é ë ê ù û ú = Þ Z 2 408 4 13 6 3 = ´ ´ Þ . . Z 2 4 = Þ Z = 2 Energy needed to remove the electron from ground state is  = + ´ = + ´ = E Z 1 2 4 ( ) 13.6 13.6 54.4 eV Hence correct option is (a). i e T e r u ev r n n n n n n = = = 2 2 p p Q u n n µ 1 and r n n µ 1 2 Þ i n n µ 1 3 Þ i i 2 1 3 1 2 = Þ i i 1 2 8 = Hence current increases 8 times correct option is (c). 5. Since five dark lines are pos si ble hence atom is ex cited to n = 6 state. The number of transition in emission line =  n n ( ) 1 2 Number of emission transition = ´ = 6 5 2 15 Hence correct option is (c). 6. A r n n = p 2 for hydrozen atom r kn n = 2 where k is constant. Þ A k n n = p 2 4 Þ A k 1 2 = p µ A A n n 1 4 = 75 Page 2 ¢ Objective Questions (Level 2) 1. F F a c = Þ GmM r mv r 2 2 = …(i) and mvr nh = 2p …(ii) From Eq. (ii) Þ v nh mr = 2p Putting this value in Eq. (i) GM r h h m r = 2 2 2 2 2 4p Þ r n h m GM = 2 2 2 2 4 p KE = = ´ = 1 2 1 2 2 2 mv m GM r GMm r PE =  GMm r Þ E = + KE PE Þ E GMm r GMm m GM n h =  =  ´ 2 4 2 2 2 2 2 p Þ E G M m n h = 2 2 2 2 3 2 2 p for ground state n = 1 Þ E G M m n =  2 2 2 2 3 2 p Hence correct option is (b). 2. We have m p n n n n n i A e T r = = ´ 2 Þ m p p n n n n n n e r u r eu r = ´ ´ ´ = 2 2 2 Þ m n e v n r n ev r n = ´ ´ ´ = ´ 1 1 2 1 1 2 2 Þ m 2 1 1 2 2 = ´ ev r and m 1 1 1 2 1 = ´ ev r Þ m m 1 2 2 = Hence magnetic moment decreases two times correct option is (b). 3. For Hlike atom E n n =  (13.6)Z eV 2 2 Here E z 2 2 4 = (13.6) and E z 1 2 =  13.6 E E 2 1  = 40.8 eV Þ 13.6 40.8 Z 2 1 1 4 ´  é ë ê ù û ú = Þ Z 2 408 4 13 6 3 = ´ ´ Þ . . Z 2 4 = Þ Z = 2 Energy needed to remove the electron from ground state is  = + ´ = + ´ = E Z 1 2 4 ( ) 13.6 13.6 54.4 eV Hence correct option is (a). i e T e r u ev r n n n n n n = = = 2 2 p p Q u n n µ 1 and r n n µ 1 2 Þ i n n µ 1 3 Þ i i 2 1 3 1 2 = Þ i i 1 2 8 = Hence current increases 8 times correct option is (c). 5. Since five dark lines are pos si ble hence atom is ex cited to n = 6 state. The number of transition in emission line =  n n ( ) 1 2 Number of emission transition = ´ = 6 5 2 15 Hence correct option is (c). 6. A r n n = p 2 for hydrozen atom r kn n = 2 where k is constant. Þ A k n n = p 2 4 Þ A k 1 2 = p µ A A n n 1 4 = 75 Taking log both sides log log A A n n 1 4 æ è ç ç ö ø ÷ ÷ = Hence it is a straight line with slope = 4 Correct option is (b) 7. For hydrogen atom i n n µ 1 3 and B i r n n n µ µ B k n n = 1 5 [ ] Q r n n ´ 2 Þ B k 2 5 2 = and B k 1 = Þ B B 2 1 5 1 2 1 32 = = Þ B B 1 2 32 = Hence magnatic field increases 32 times. The correct option is (d). 8. For Hatom Lyman se ries is given by 1 1 1 1 2 2 l =  é ë ê ù û ú R n for first line n = 2 Þ 1 1 1 2 2 l =  é ë ê ù û ú R Þ 1 3 4 l = R Momentum of photon P h p = l Let momentum of atom p A Q Initial momentum was zero. Hence using momentum conservation law, we get p p A B = Þ Mv h hR = = l 3 4 Þ v hR M = 3 4 Hence the correct option is (a). 9. Light wave equation is 200 10 15 1 V/m 1.5 sec sin ( ) ´  t ´ ´  cos ( ) 0.5 sec 10 15 1 t Here maximum frequency = ´ 1.5 10 2 15 p Maxmum incident energy = ´ ´ ´ ´   1.5 6.6 1.6 10 2 10 10 15 34 19 p = 0.98 eV Since work function = > 2eV maximum energy hence no emission of electrons. Thus correct option is (d). 10. Since in Balmer se ries of Hlike atom wave lengths (in vis i ble re gion) are found same or smaller hence the gas was ini tially in sec ond ex cited state. Cor rect op tion is (c). 11. For Hatom T n H = 2 3 p and for Hlike atom T n z x = 2 3 2 p For Hatom in ground state T H = 2p For Hlike atom in first excited state T z z x = ´ = ´ ( ) 2 2 2 8 3 2 2 p p But T T H x = 2 Þ 2 2 2 8 2 p p = ´ ´ z Þ z 2 16 = Þ z = 4 Hence correct option is (c). 12. For K a line of Xray 1 1 2 2 l =  a c z ( ) Q z (atomic No.) for Pb Pb Pb 204 206 208 , , are same hence l l l 1 2 3 = = . Hence correct option is (c). 13. The cor rect op tion is (d). 14. Since E n n = 13.6 eV 2 Þ E 1 =  13.6 eV and first excited state E 2 4 = 13.6 eV 76 Page 3 ¢ Objective Questions (Level 2) 1. F F a c = Þ GmM r mv r 2 2 = …(i) and mvr nh = 2p …(ii) From Eq. (ii) Þ v nh mr = 2p Putting this value in Eq. (i) GM r h h m r = 2 2 2 2 2 4p Þ r n h m GM = 2 2 2 2 4 p KE = = ´ = 1 2 1 2 2 2 mv m GM r GMm r PE =  GMm r Þ E = + KE PE Þ E GMm r GMm m GM n h =  =  ´ 2 4 2 2 2 2 2 p Þ E G M m n h = 2 2 2 2 3 2 2 p for ground state n = 1 Þ E G M m n =  2 2 2 2 3 2 p Hence correct option is (b). 2. We have m p n n n n n i A e T r = = ´ 2 Þ m p p n n n n n n e r u r eu r = ´ ´ ´ = 2 2 2 Þ m n e v n r n ev r n = ´ ´ ´ = ´ 1 1 2 1 1 2 2 Þ m 2 1 1 2 2 = ´ ev r and m 1 1 1 2 1 = ´ ev r Þ m m 1 2 2 = Hence magnetic moment decreases two times correct option is (b). 3. For Hlike atom E n n =  (13.6)Z eV 2 2 Here E z 2 2 4 = (13.6) and E z 1 2 =  13.6 E E 2 1  = 40.8 eV Þ 13.6 40.8 Z 2 1 1 4 ´  é ë ê ù û ú = Þ Z 2 408 4 13 6 3 = ´ ´ Þ . . Z 2 4 = Þ Z = 2 Energy needed to remove the electron from ground state is  = + ´ = + ´ = E Z 1 2 4 ( ) 13.6 13.6 54.4 eV Hence correct option is (a). i e T e r u ev r n n n n n n = = = 2 2 p p Q u n n µ 1 and r n n µ 1 2 Þ i n n µ 1 3 Þ i i 2 1 3 1 2 = Þ i i 1 2 8 = Hence current increases 8 times correct option is (c). 5. Since five dark lines are pos si ble hence atom is ex cited to n = 6 state. The number of transition in emission line =  n n ( ) 1 2 Number of emission transition = ´ = 6 5 2 15 Hence correct option is (c). 6. A r n n = p 2 for hydrozen atom r kn n = 2 where k is constant. Þ A k n n = p 2 4 Þ A k 1 2 = p µ A A n n 1 4 = 75 Taking log both sides log log A A n n 1 4 æ è ç ç ö ø ÷ ÷ = Hence it is a straight line with slope = 4 Correct option is (b) 7. For hydrogen atom i n n µ 1 3 and B i r n n n µ µ B k n n = 1 5 [ ] Q r n n ´ 2 Þ B k 2 5 2 = and B k 1 = Þ B B 2 1 5 1 2 1 32 = = Þ B B 1 2 32 = Hence magnatic field increases 32 times. The correct option is (d). 8. For Hatom Lyman se ries is given by 1 1 1 1 2 2 l =  é ë ê ù û ú R n for first line n = 2 Þ 1 1 1 2 2 l =  é ë ê ù û ú R Þ 1 3 4 l = R Momentum of photon P h p = l Let momentum of atom p A Q Initial momentum was zero. Hence using momentum conservation law, we get p p A B = Þ Mv h hR = = l 3 4 Þ v hR M = 3 4 Hence the correct option is (a). 9. Light wave equation is 200 10 15 1 V/m 1.5 sec sin ( ) ´  t ´ ´  cos ( ) 0.5 sec 10 15 1 t Here maximum frequency = ´ 1.5 10 2 15 p Maxmum incident energy = ´ ´ ´ ´   1.5 6.6 1.6 10 2 10 10 15 34 19 p = 0.98 eV Since work function = > 2eV maximum energy hence no emission of electrons. Thus correct option is (d). 10. Since in Balmer se ries of Hlike atom wave lengths (in vis i ble re gion) are found same or smaller hence the gas was ini tially in sec ond ex cited state. Cor rect op tion is (c). 11. For Hatom T n H = 2 3 p and for Hlike atom T n z x = 2 3 2 p For Hatom in ground state T H = 2p For Hlike atom in first excited state T z z x = ´ = ´ ( ) 2 2 2 8 3 2 2 p p But T T H x = 2 Þ 2 2 2 8 2 p p = ´ ´ z Þ z 2 16 = Þ z = 4 Hence correct option is (c). 12. For K a line of Xray 1 1 2 2 l =  a c z ( ) Q z (atomic No.) for Pb Pb Pb 204 206 208 , , are same hence l l l 1 2 3 = = . Hence correct option is (c). 13. The cor rect op tion is (d). 14. Since E n n = 13.6 eV 2 Þ E 1 =  13.6 eV and first excited state E 2 4 = 13.6 eV 76 Þ E 2 =  3.4 eV DE E E =  = 2 1 10.2 if K < 10.2 eV The electron collide elastically with Hatom in ground state. The correct option is (c). 15. For Lyman se ries 1 1 1 1 2 2 l =  é ë ê ù û ú R n here n = 3 Þ 1 1 1 9 l =  é ë ê ù û ú R Þ 1 8 9 l = R P h Rh photon = = l 8 9 But P P Photon Hatom = Þ 8 9 Rh M v p = ´ Þ v = ´ ´ ´ ´ ´ ´ ´   8 10 10 9 10 7 34 31 1.097 6.6 1837 9.1 Þ v = 4 m/s Hence correct option is (a). 16. Power = = ´ ´ ´ =  VI 150 10 10 10 1500 3 3 W The 99% power heated the target hence Heating power = ´ = ´ 99 100 1500 15 99 W The rate at which target is heated per sec. (in cal) = ´  15 99 355 4.2 ~ Q1 1 J 4.2 cal = é ë ê ù û ú Hence correct option is (c). 17. E z n n =  ´ 13.6 eV 2 2 E z 3 2 9 =  13.6 eV and E z 4 2 16 = 13.6 eV DE E E z =  = ´  é ë ê ù û ú 4 3 2 1 9 1 16 ( ) 13.6 eV Þ DE = ´ ´ ´ = 13.6 eV) 32.4 eV ( 2 7 16 9 2 Þ z 2 9 7 49 = ´ ´ ´ = 16 32.4 13.6 Þ z = 7 Hence correct option is (d). 18. l = = = h p h m K h m eV p p 2 2 Þ 10 10 2 1836 10 10 13 34 31 19     = ´ ´ ´ ´ ´ ´ ´ 6.6 9.1 1.6 V Þ V = ´ 8.15 volt 10 4 Hence correct option is (b). 19. Since E n n µ 1 2 and L n n µ Hence E L n n µ 1 2 The correct option is (d). 20. Since m p n n n eu r = 2 Q u n n µ 1 and r n n µ 2 Þ m n kn = Where k is constant for Hatom For ground state m 1 1 = ´ = k k …(i) For third excited state n = 4 m 2 4 4 = ´ = k k …(ii) From Eqs. (i) and (ii) we get m m 2 1 4 = Hence correct option is (d). 21. By con ser v a tion of mo men tum M v M M v H H H = + ¢ ( ) Þ v v ¢ = 2 Let initial KE of Hatom = K Final KE of eachHatom = K 2 For excitation K E E 2 4 2 1 =  =  + 13.6 13.6 Þ K 2 = 10.2 eV Þ K = 20.4 eV Hence correct option is (a). 77 Page 4 ¢ Objective Questions (Level 2) 1. F F a c = Þ GmM r mv r 2 2 = …(i) and mvr nh = 2p …(ii) From Eq. (ii) Þ v nh mr = 2p Putting this value in Eq. (i) GM r h h m r = 2 2 2 2 2 4p Þ r n h m GM = 2 2 2 2 4 p KE = = ´ = 1 2 1 2 2 2 mv m GM r GMm r PE =  GMm r Þ E = + KE PE Þ E GMm r GMm m GM n h =  =  ´ 2 4 2 2 2 2 2 p Þ E G M m n h = 2 2 2 2 3 2 2 p for ground state n = 1 Þ E G M m n =  2 2 2 2 3 2 p Hence correct option is (b). 2. We have m p n n n n n i A e T r = = ´ 2 Þ m p p n n n n n n e r u r eu r = ´ ´ ´ = 2 2 2 Þ m n e v n r n ev r n = ´ ´ ´ = ´ 1 1 2 1 1 2 2 Þ m 2 1 1 2 2 = ´ ev r and m 1 1 1 2 1 = ´ ev r Þ m m 1 2 2 = Hence magnetic moment decreases two times correct option is (b). 3. For Hlike atom E n n =  (13.6)Z eV 2 2 Here E z 2 2 4 = (13.6) and E z 1 2 =  13.6 E E 2 1  = 40.8 eV Þ 13.6 40.8 Z 2 1 1 4 ´  é ë ê ù û ú = Þ Z 2 408 4 13 6 3 = ´ ´ Þ . . Z 2 4 = Þ Z = 2 Energy needed to remove the electron from ground state is  = + ´ = + ´ = E Z 1 2 4 ( ) 13.6 13.6 54.4 eV Hence correct option is (a). i e T e r u ev r n n n n n n = = = 2 2 p p Q u n n µ 1 and r n n µ 1 2 Þ i n n µ 1 3 Þ i i 2 1 3 1 2 = Þ i i 1 2 8 = Hence current increases 8 times correct option is (c). 5. Since five dark lines are pos si ble hence atom is ex cited to n = 6 state. The number of transition in emission line =  n n ( ) 1 2 Number of emission transition = ´ = 6 5 2 15 Hence correct option is (c). 6. A r n n = p 2 for hydrozen atom r kn n = 2 where k is constant. Þ A k n n = p 2 4 Þ A k 1 2 = p µ A A n n 1 4 = 75 Taking log both sides log log A A n n 1 4 æ è ç ç ö ø ÷ ÷ = Hence it is a straight line with slope = 4 Correct option is (b) 7. For hydrogen atom i n n µ 1 3 and B i r n n n µ µ B k n n = 1 5 [ ] Q r n n ´ 2 Þ B k 2 5 2 = and B k 1 = Þ B B 2 1 5 1 2 1 32 = = Þ B B 1 2 32 = Hence magnatic field increases 32 times. The correct option is (d). 8. For Hatom Lyman se ries is given by 1 1 1 1 2 2 l =  é ë ê ù û ú R n for first line n = 2 Þ 1 1 1 2 2 l =  é ë ê ù û ú R Þ 1 3 4 l = R Momentum of photon P h p = l Let momentum of atom p A Q Initial momentum was zero. Hence using momentum conservation law, we get p p A B = Þ Mv h hR = = l 3 4 Þ v hR M = 3 4 Hence the correct option is (a). 9. Light wave equation is 200 10 15 1 V/m 1.5 sec sin ( ) ´  t ´ ´  cos ( ) 0.5 sec 10 15 1 t Here maximum frequency = ´ 1.5 10 2 15 p Maxmum incident energy = ´ ´ ´ ´   1.5 6.6 1.6 10 2 10 10 15 34 19 p = 0.98 eV Since work function = > 2eV maximum energy hence no emission of electrons. Thus correct option is (d). 10. Since in Balmer se ries of Hlike atom wave lengths (in vis i ble re gion) are found same or smaller hence the gas was ini tially in sec ond ex cited state. Cor rect op tion is (c). 11. For Hatom T n H = 2 3 p and for Hlike atom T n z x = 2 3 2 p For Hatom in ground state T H = 2p For Hlike atom in first excited state T z z x = ´ = ´ ( ) 2 2 2 8 3 2 2 p p But T T H x = 2 Þ 2 2 2 8 2 p p = ´ ´ z Þ z 2 16 = Þ z = 4 Hence correct option is (c). 12. For K a line of Xray 1 1 2 2 l =  a c z ( ) Q z (atomic No.) for Pb Pb Pb 204 206 208 , , are same hence l l l 1 2 3 = = . Hence correct option is (c). 13. The cor rect op tion is (d). 14. Since E n n = 13.6 eV 2 Þ E 1 =  13.6 eV and first excited state E 2 4 = 13.6 eV 76 Þ E 2 =  3.4 eV DE E E =  = 2 1 10.2 if K < 10.2 eV The electron collide elastically with Hatom in ground state. The correct option is (c). 15. For Lyman se ries 1 1 1 1 2 2 l =  é ë ê ù û ú R n here n = 3 Þ 1 1 1 9 l =  é ë ê ù û ú R Þ 1 8 9 l = R P h Rh photon = = l 8 9 But P P Photon Hatom = Þ 8 9 Rh M v p = ´ Þ v = ´ ´ ´ ´ ´ ´ ´   8 10 10 9 10 7 34 31 1.097 6.6 1837 9.1 Þ v = 4 m/s Hence correct option is (a). 16. Power = = ´ ´ ´ =  VI 150 10 10 10 1500 3 3 W The 99% power heated the target hence Heating power = ´ = ´ 99 100 1500 15 99 W The rate at which target is heated per sec. (in cal) = ´  15 99 355 4.2 ~ Q1 1 J 4.2 cal = é ë ê ù û ú Hence correct option is (c). 17. E z n n =  ´ 13.6 eV 2 2 E z 3 2 9 =  13.6 eV and E z 4 2 16 = 13.6 eV DE E E z =  = ´  é ë ê ù û ú 4 3 2 1 9 1 16 ( ) 13.6 eV Þ DE = ´ ´ ´ = 13.6 eV) 32.4 eV ( 2 7 16 9 2 Þ z 2 9 7 49 = ´ ´ ´ = 16 32.4 13.6 Þ z = 7 Hence correct option is (d). 18. l = = = h p h m K h m eV p p 2 2 Þ 10 10 2 1836 10 10 13 34 31 19     = ´ ´ ´ ´ ´ ´ ´ 6.6 9.1 1.6 V Þ V = ´ 8.15 volt 10 4 Hence correct option is (b). 19. Since E n n µ 1 2 and L n n µ Hence E L n n µ 1 2 The correct option is (d). 20. Since m p n n n eu r = 2 Q u n n µ 1 and r n n µ 2 Þ m n kn = Where k is constant for Hatom For ground state m 1 1 = ´ = k k …(i) For third excited state n = 4 m 2 4 4 = ´ = k k …(ii) From Eqs. (i) and (ii) we get m m 2 1 4 = Hence correct option is (d). 21. By con ser v a tion of mo men tum M v M M v H H H = + ¢ ( ) Þ v v ¢ = 2 Let initial KE of Hatom = K Final KE of eachHatom = K 2 For excitation K E E 2 4 2 1 =  =  + 13.6 13.6 Þ K 2 = 10.2 eV Þ K = 20.4 eV Hence correct option is (a). 77 22. We know that for Hlike atom E K n n =  Þ K n = 3.4 eV l = = h p h m K e 2 = ´ ´ ´ ´ ´ ´    6.6 9.1 3.4 1.6 10 2 10 10 34 31 19 Þ l = 6.6 Å Hence both options (a) and (b) are correct. Hence answer is (c) both are correct. 2325. 0.6e hc W = ´   4950 10 10 …(i) 1.1 e hc W =  l 2 …(ii) Subtracting Eq. (i) from Eq. (ii) we get 0.5 e hc =  ´ é ë ê ù û ú  1 1 4950 10 2 10 l Þ 0.5 1.6 6.62 ´ ´ ´ ´ ´ + ´ =    10 10 3 10 1 4950 10 1 19 34 8 10 2 l Þ l 2 ~  4111 Å From Eq. (i) W = ´ ´ ´ ´ ´ ´     6.6 4.95 1.6 1.9 eV 10 3 10 10 10 34 8 7 19 ~ 23. W = 1.9 eV Hence correct option is (c). 24. l = 4111 Å Hence correct option is (c) 25. Since magnatic field does not change the KE of elec trons hence re tard ing potential re main same. Hence correct option is (c). 26. ( ) max KE eV eV eV =  = 5 3 2 = ´ ´  2 19 1.6 J l min = = h p h m K e 2 Þ l min = ´ ´ ´ ´ ´ ´    6.6 2 9.1 1.6 10 10 2 10 34 31 19 = 8.69 Å Hence correct option is (b). 27. Photo emis sion will stop when potential of sphere be comes stop ping po ten tial Þ 1 4 2 0 p e q r = V Since ( ) max KE eV = 0 hence V 0 2 = Þ q r = 8 0 pe coulomb Hence correct option is (b) 28. Let t be the time for photo emis sion 1 4 2 0 pe ´ ´ = q r t t r q = 8 0 pe Intensity of light at 0.8 from source I = ´ ´  ´   3.2 0.8 W/m 10 4 4 10 3 2 4 2 p ( ) ~ Energy incident on the sphere in unit time E 1 3 2 4 8 8 10 4 10 10 = ´ ´ ´ ´ = ´    p ( ) 8.04 W Energy of each photon E 2 19 19 5 10 8 10 = ´ ´ = ´   1.6 J Total number of photons incident on the sphere per second h = = ´ ´ =   E E 1 2 8 19 11 10 8 10 10 8.04 Since 10 6 photons emit one electron. Hence the total number of photoelectron per sec is n n 2 1 1 11 6 5 10 10 10 10 = = = Therefore, q n e t t = ´ ´ = ´ ´ ´  2 5 19 10 10 1.6 Þ 9 10 10 2 9 14 ´ ´ ´ ´ =  1.6 t r 78 Page 5 ¢ Objective Questions (Level 2) 1. F F a c = Þ GmM r mv r 2 2 = …(i) and mvr nh = 2p …(ii) From Eq. (ii) Þ v nh mr = 2p Putting this value in Eq. (i) GM r h h m r = 2 2 2 2 2 4p Þ r n h m GM = 2 2 2 2 4 p KE = = ´ = 1 2 1 2 2 2 mv m GM r GMm r PE =  GMm r Þ E = + KE PE Þ E GMm r GMm m GM n h =  =  ´ 2 4 2 2 2 2 2 p Þ E G M m n h = 2 2 2 2 3 2 2 p for ground state n = 1 Þ E G M m n =  2 2 2 2 3 2 p Hence correct option is (b). 2. We have m p n n n n n i A e T r = = ´ 2 Þ m p p n n n n n n e r u r eu r = ´ ´ ´ = 2 2 2 Þ m n e v n r n ev r n = ´ ´ ´ = ´ 1 1 2 1 1 2 2 Þ m 2 1 1 2 2 = ´ ev r and m 1 1 1 2 1 = ´ ev r Þ m m 1 2 2 = Hence magnetic moment decreases two times correct option is (b). 3. For Hlike atom E n n =  (13.6)Z eV 2 2 Here E z 2 2 4 = (13.6) and E z 1 2 =  13.6 E E 2 1  = 40.8 eV Þ 13.6 40.8 Z 2 1 1 4 ´  é ë ê ù û ú = Þ Z 2 408 4 13 6 3 = ´ ´ Þ . . Z 2 4 = Þ Z = 2 Energy needed to remove the electron from ground state is  = + ´ = + ´ = E Z 1 2 4 ( ) 13.6 13.6 54.4 eV Hence correct option is (a). i e T e r u ev r n n n n n n = = = 2 2 p p Q u n n µ 1 and r n n µ 1 2 Þ i n n µ 1 3 Þ i i 2 1 3 1 2 = Þ i i 1 2 8 = Hence current increases 8 times correct option is (c). 5. Since five dark lines are pos si ble hence atom is ex cited to n = 6 state. The number of transition in emission line =  n n ( ) 1 2 Number of emission transition = ´ = 6 5 2 15 Hence correct option is (c). 6. A r n n = p 2 for hydrozen atom r kn n = 2 where k is constant. Þ A k n n = p 2 4 Þ A k 1 2 = p µ A A n n 1 4 = 75 Taking log both sides log log A A n n 1 4 æ è ç ç ö ø ÷ ÷ = Hence it is a straight line with slope = 4 Correct option is (b) 7. For hydrogen atom i n n µ 1 3 and B i r n n n µ µ B k n n = 1 5 [ ] Q r n n ´ 2 Þ B k 2 5 2 = and B k 1 = Þ B B 2 1 5 1 2 1 32 = = Þ B B 1 2 32 = Hence magnatic field increases 32 times. The correct option is (d). 8. For Hatom Lyman se ries is given by 1 1 1 1 2 2 l =  é ë ê ù û ú R n for first line n = 2 Þ 1 1 1 2 2 l =  é ë ê ù û ú R Þ 1 3 4 l = R Momentum of photon P h p = l Let momentum of atom p A Q Initial momentum was zero. Hence using momentum conservation law, we get p p A B = Þ Mv h hR = = l 3 4 Þ v hR M = 3 4 Hence the correct option is (a). 9. Light wave equation is 200 10 15 1 V/m 1.5 sec sin ( ) ´  t ´ ´  cos ( ) 0.5 sec 10 15 1 t Here maximum frequency = ´ 1.5 10 2 15 p Maxmum incident energy = ´ ´ ´ ´   1.5 6.6 1.6 10 2 10 10 15 34 19 p = 0.98 eV Since work function = > 2eV maximum energy hence no emission of electrons. Thus correct option is (d). 10. Since in Balmer se ries of Hlike atom wave lengths (in vis i ble re gion) are found same or smaller hence the gas was ini tially in sec ond ex cited state. Cor rect op tion is (c). 11. For Hatom T n H = 2 3 p and for Hlike atom T n z x = 2 3 2 p For Hatom in ground state T H = 2p For Hlike atom in first excited state T z z x = ´ = ´ ( ) 2 2 2 8 3 2 2 p p But T T H x = 2 Þ 2 2 2 8 2 p p = ´ ´ z Þ z 2 16 = Þ z = 4 Hence correct option is (c). 12. For K a line of Xray 1 1 2 2 l =  a c z ( ) Q z (atomic No.) for Pb Pb Pb 204 206 208 , , are same hence l l l 1 2 3 = = . Hence correct option is (c). 13. The cor rect op tion is (d). 14. Since E n n = 13.6 eV 2 Þ E 1 =  13.6 eV and first excited state E 2 4 = 13.6 eV 76 Þ E 2 =  3.4 eV DE E E =  = 2 1 10.2 if K < 10.2 eV The electron collide elastically with Hatom in ground state. The correct option is (c). 15. For Lyman se ries 1 1 1 1 2 2 l =  é ë ê ù û ú R n here n = 3 Þ 1 1 1 9 l =  é ë ê ù û ú R Þ 1 8 9 l = R P h Rh photon = = l 8 9 But P P Photon Hatom = Þ 8 9 Rh M v p = ´ Þ v = ´ ´ ´ ´ ´ ´ ´   8 10 10 9 10 7 34 31 1.097 6.6 1837 9.1 Þ v = 4 m/s Hence correct option is (a). 16. Power = = ´ ´ ´ =  VI 150 10 10 10 1500 3 3 W The 99% power heated the target hence Heating power = ´ = ´ 99 100 1500 15 99 W The rate at which target is heated per sec. (in cal) = ´  15 99 355 4.2 ~ Q1 1 J 4.2 cal = é ë ê ù û ú Hence correct option is (c). 17. E z n n =  ´ 13.6 eV 2 2 E z 3 2 9 =  13.6 eV and E z 4 2 16 = 13.6 eV DE E E z =  = ´  é ë ê ù û ú 4 3 2 1 9 1 16 ( ) 13.6 eV Þ DE = ´ ´ ´ = 13.6 eV) 32.4 eV ( 2 7 16 9 2 Þ z 2 9 7 49 = ´ ´ ´ = 16 32.4 13.6 Þ z = 7 Hence correct option is (d). 18. l = = = h p h m K h m eV p p 2 2 Þ 10 10 2 1836 10 10 13 34 31 19     = ´ ´ ´ ´ ´ ´ ´ 6.6 9.1 1.6 V Þ V = ´ 8.15 volt 10 4 Hence correct option is (b). 19. Since E n n µ 1 2 and L n n µ Hence E L n n µ 1 2 The correct option is (d). 20. Since m p n n n eu r = 2 Q u n n µ 1 and r n n µ 2 Þ m n kn = Where k is constant for Hatom For ground state m 1 1 = ´ = k k …(i) For third excited state n = 4 m 2 4 4 = ´ = k k …(ii) From Eqs. (i) and (ii) we get m m 2 1 4 = Hence correct option is (d). 21. By con ser v a tion of mo men tum M v M M v H H H = + ¢ ( ) Þ v v ¢ = 2 Let initial KE of Hatom = K Final KE of eachHatom = K 2 For excitation K E E 2 4 2 1 =  =  + 13.6 13.6 Þ K 2 = 10.2 eV Þ K = 20.4 eV Hence correct option is (a). 77 22. We know that for Hlike atom E K n n =  Þ K n = 3.4 eV l = = h p h m K e 2 = ´ ´ ´ ´ ´ ´    6.6 9.1 3.4 1.6 10 2 10 10 34 31 19 Þ l = 6.6 Å Hence both options (a) and (b) are correct. Hence answer is (c) both are correct. 2325. 0.6e hc W = ´   4950 10 10 …(i) 1.1 e hc W =  l 2 …(ii) Subtracting Eq. (i) from Eq. (ii) we get 0.5 e hc =  ´ é ë ê ù û ú  1 1 4950 10 2 10 l Þ 0.5 1.6 6.62 ´ ´ ´ ´ ´ + ´ =    10 10 3 10 1 4950 10 1 19 34 8 10 2 l Þ l 2 ~  4111 Å From Eq. (i) W = ´ ´ ´ ´ ´ ´     6.6 4.95 1.6 1.9 eV 10 3 10 10 10 34 8 7 19 ~ 23. W = 1.9 eV Hence correct option is (c). 24. l = 4111 Å Hence correct option is (c) 25. Since magnatic field does not change the KE of elec trons hence re tard ing potential re main same. Hence correct option is (c). 26. ( ) max KE eV eV eV =  = 5 3 2 = ´ ´  2 19 1.6 J l min = = h p h m K e 2 Þ l min = ´ ´ ´ ´ ´ ´    6.6 2 9.1 1.6 10 10 2 10 34 31 19 = 8.69 Å Hence correct option is (b). 27. Photo emis sion will stop when potential of sphere be comes stop ping po ten tial Þ 1 4 2 0 p e q r = V Since ( ) max KE eV = 0 hence V 0 2 = Þ q r = 8 0 pe coulomb Hence correct option is (b) 28. Let t be the time for photo emis sion 1 4 2 0 pe ´ ´ = q r t t r q = 8 0 pe Intensity of light at 0.8 from source I = ´ ´  ´   3.2 0.8 W/m 10 4 4 10 3 2 4 2 p ( ) ~ Energy incident on the sphere in unit time E 1 3 2 4 8 8 10 4 10 10 = ´ ´ ´ ´ = ´    p ( ) 8.04 W Energy of each photon E 2 19 19 5 10 8 10 = ´ ´ = ´   1.6 J Total number of photons incident on the sphere per second h = = ´ ´ =   E E 1 2 8 19 11 10 8 10 10 8.04 Since 10 6 photons emit one electron. Hence the total number of photoelectron per sec is n n 2 1 1 11 6 5 10 10 10 10 = = = Therefore, q n e t t = ´ ´ = ´ ´ ´  2 5 19 10 10 1.6 Þ 9 10 10 2 9 14 ´ ´ ´ ´ =  1.6 t r 78 t = ´ ´ ´ ´   2 8 10 9 10 3 5 1.6 Þ t = 111 s Hence correct option is (c). ¢ More than one op tions are cor rect 1. Since l 0 = hc eV if v in creases l 0 de creases hence the in ter val be tween l a K and l 0 as well as l b K and l 0 in creases. The correct options are (b) and (c). 2. R n µ 2 , V n µ 1 and E n µ 1 2 for Bohr model of Hatom Þ VR n µ and V E n µ Hence, the correct options are (a) and (c). 3. For Bohr model of Hatom L n µ , r n µ 2 and T n µ 3 Hence rL T is independent of n L T n µ 1 2 and T r n L n µ µ , 3 Hence correct options are (a), (b) and (c). 4. Q l = h mv and l = h mK 2 Hence heavy particle has smallest wavelength when speed and KE both particle are same. The correct options are (a) and (c). 5. Since there are six different wave length Hence, final state will be n = 4. Since two wavelengths are longer than l 0 [(From n = ® 4 3 and n = ® 3 2)] Hence initial state was n = 2 and there are three transitions shown as (1), (2) and (3) belonging to Lymen series. Hence correct options are (a), (b) and (d). 6. Q f a z b =  ( ) f versus z is a straight line Q f c = l Þ 1 l =  a c z b ( ) hence 1 l versus z is a straight line f a z b =  2 ( ) Þ log log log ( ) f a z b = +  2 which is a straight line Hence correct options are (a), (b) and (c). ¢ Match the Columns : 1. Lymen se ries lies in UV re gion, Balmer se ries lies in vis i ble re gion and Paschen and Brackelt se ries lie in infrared re gion. Hence (a) r (b) q (c) p (d) p 2. For Hatom E n n = 13.6 eV 2 Þ E 2 2 2 = 13.6 eV = 13.6 eV 4 Ionization energy from first excited state of Hatom E E =  = 2 4 13.6 eV …(i) For He + ion E n Z H ( ( ) He) 13.6 eV =  ´ 2 2 for He + = z 2 Þ E n ( ) He =  ´ 13.6 eV n 2 4 Ionization energy of He + atom from ground state = ´ = ´ ( ) 13.6 eV 4 4 4 E from Eq. (i) = 16 E 79 n = 4 n = 3 n = 2 n = 1 1 2 3Read More

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