Tiêu đề DPP-5 SOLUTION.pdf ✅

CLASS : XIth SUBJECT : CHEMISTRY DATE : DPP No. : 5 1 (c) Total no. of protons in all the elements from at. no. 1 to at no. n = n × (n + 1)/2. 2 (b) Frequency (n) = 1 time period (T) Here, T = 5 × 10―3 s n = 1 5 × 10―3 = 0.2 × 103 = 2 × 102 s ―1 3 (a) e m for : (i) neutron = 0 1 = 0 (ii)α-particle = 2 4 = 0.5 (iii)proton = 1 1 = 1 (iv)electron = 1 1/1837 = 1837 4 (d) It is the definition of degenerate orbitals. 5 (a) N and P have 3 unpaired electrons in 2p and 3p respectively; V has 3 unpaired electrons in 3d . 6 (a) Momentum of photon = mu = h λ ( ∵ λ = h mu) = 6.6 × 10―34 2 × 10―11 = 3.3 × 10―23kg m s ―1 7 (c) 35 = 1s 2 ,2s 2 ,2p 6 ,3s 2 ,3p 6 ,4s 2 ,3d 10,4p 5 Thus, it contains 7 electrons in 4th or outermost shell 8 (b) Follow Dalton’s assumptions. 9 (d) Schrödinger proposed the concept of orbitals –a three-dimensional region in which Topic :- STRUCTURE OF ATOM Solutions
probability for finding electron is maximum. 10 (d) All are facts 11 (d) Pb sheets cut X-rays. 12 (c) Maximum no. of electron in an orbit = 2n 2 . 13 (c) Total values of ′m′ in a given shell = n 2 . 14 (d) 1 λ = Z 2 , RH[ 1 n 2 1 ― 1 n 2 2 ] For He +, 1 λ = 2 2 .RH[ 1 2 2 ― 1 4 2] = 4 × 3 16 = 3 4 For H, 1 λ = 1 2 .RH[ 1 1 2 ― 1 2 2] = 3 4 Hence, for hydrogen n = 2 to n = 1. 15 (b) After filling up of electron in np, the next electron occupies (n + 1)s level. 16 (c) 1 λLyman = RH[ 1 1 2 ― 1 ∞2]; 1 λBalmer = RH[ 1 2 2 ― 1 ∞2] 17 (c) Work function for Cs is minimum. 18 (c) It is famous Schrödinger wave equation. 19 (a) Tritium has only one electron. 20 (b) A characteristic of cathode rays particles (electrons).
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. C B A D A A C B D D Q. 11 12 13 14 15 16 17 18 19 20 A. D C C D B C C C A B