Tiêu đề DPP-10 SOLUTION.pdf ✅

CLASS : XIth SUBJECT : CHEMISTRY DATE : DPP No. : 10 1 (d) The values of quantum number will give idea about the last subshell of element. From that value we can find the atomic number of element, n = 3 means 3rd-shell l = 0 m = 0 }means subshell It means it is 3s-subshell which can have 1 or 2 electrons. ∴ Configuration of element is 1s 2 ,2s 2 ,2p 6 ,3s 1―2 ∴ Atomic i.e., number is 11 or 12. 2 (a) hv = work function +KE; or hv = hv0 + KE; hv0 = work function = hc λ0 ; where λ0 is threshold wavelength. 3 (a) The Sc atom has 3d 1 ,4s 2 configuration. 4 (a) Wave number of spectral line in emission spectrum of hydrogen, v = RH( 1 n 2 1 ― 1 n 2 2 ) ...(i) Given, v = 8 9 RH On putting the value of v in Eq. (i), we get 8 9 = RH( 1 n 2 1 ― 1 n 2 2 ) 8 9 = 1 (1) 2 ― 1 n 2 2 8 9 ― 1 = ― 1 n 2 2 1 3 = 1 n2 Topic :- STRUCTURE OF ATOM Solutions
∴ n2 = 3 Hence, electron jumps from n2 = 3 to n1 = 1 5 (b) J.J. Thomson (1987) was first experimentally demonstrated particle nature of electron. It was first of all proposed by Millikan’s oil drop experiment. 6 (b) Angular momentum for n and (n + 1) shells are nh 2π and (n + 1) h 2π. 7 (b) The volume of nucleus : volume of atom, 4 3 πr 3 n: 4 3 πr 3 atom. 8 (c) O 2― has 10 electrons but 8 neutrons ( 8O 16). 10 (c) Possible mol. wt. may be 18,20,19,20,22,21 respectively for H 1H 1O 16 ,H 2H 2O 16 ,H 1H 2O 16 ,H 1 H 1O 18 ,H 2H 2O 18 ,H 1H 2O 18 . 11 (c) Magnetic moment = [n(n + 2)] where n is number of unpaired electrons . 12 (d) Hertz for the first time noticed the effect. 13 (b) Cr (24):[Ar]3d 5 4s 1 Cr 3+:[Ar]3d 3 4s 0 14 (d) A part of energy of photon (hv-work function) is used for kinetic energy of electrons. 15 (b) e m for electron (e) = 1.6 × 10―19 9.1 × 10―28 = 1.758 × 108 e m for proton (p) = 1.6 × 10―19 1.672 × 10―24 = 9.56 × 104 e m for neutron (n) = 0 1.675 × 10―24 = 0 e m for α ― particle = 2 4 = 0.5 Hence, the increasing order of e m is as n < α < p < e 16 (d) Ionisation energy of nitrogen =energy of photon = Nh c λ where, N = 6.02 × 1023 c = 3 × 108ms ―1
λ = 854 Å = 854 × 10―10m = 6.02 × 1023 × 6.6 × 10―34 × 3 × 108 854 × 10―10 = 1.4 × 106 J mol ―1 = 1.4 × 103 kJ mol ―1 17 (a) e/m for proton = 1 1 ;e/m for α = 2 4 18 (a) E = n hc λ h = 6.6 × 10―34 Js or 1J = n × 6.6 × 10―34 × 3 × 108 4000 × 10―10 19 (c) We know that the energy is emitted in the form of quanta and is given by, ∆E = hv = hc λ or λ = hc ∆E = 6.62 × 10―27 × 3 × 1010 3 × 1.6 × 10―12 = 4.14 × 10―5 cm = 4140 Å
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. D A A A B B B C D C Q. 11 12 13 14 15 16 17 18 19 20 A. C D B D B D A A C B