Question

2. Determine the corresponding emission wavelength (in nm) for the emission energies (in kJ/mol) ot each Group IA or IIA metal given in the table below. Element Emission Energy (kJ/mol) Wavelength (nm) Li 178.2 Na 203.5 К 155.3 156.5 184.8 198.7 Са 191.4 216.1 Sr 178.2 197.5 Ва 144.5 161.9

Answer 1

DaAlready we know that, Eche where E= energy X = Wavelengti c = light velocity = 3x108m. sed ha planches constant 76.625x1034 Judec: Givers li has Energy = 178.2 IcJln ke. I mole having 6.023 61023x10 3 -178.2 kJ latom - ? » 178.2 = 29.586X16?PKJ latom. 6.023x1023 -23 ii Ech 6.625 x103 J x 3x108m.se =ne = 29.586x10 23 X5 . = 6.625 x 3 x 103478 ('T KJ=181) 29.586 x103 x1623 = 0.67177 x1őom finom - jong = 0.67177x166x10 nm) -im = 0.67177x102nn saam = 109nm) = 671.771

6 Na has €2.6 3.5 leilmol 6.023x1023 atomos - 203.5 Tatom => 203,5 = 33.787 X10 KT 6.023x1023 .. E= he ^ = 6.625*10*3.896 x3x108 museet 33.787 X102103 ] = 6.625x3 34 +8+23-3 33.787 0,5 8 824 X 10° ) -0.58824x10610 In = 588.24 nm Ő K har E = 155.3 kJlmol. latonk, 155.3 -23 = 25.784 X10T 6.023x103 E=he 8 34 6.625 XIO X 3x10 1 Asec m.seat 25.784 x 10²3 10² J = 0.7708 vios ir = 770.8 om rii) k hou E = 156.5 - latom has (Il mol 156.5 . -2.3 = 25.9837X1D KJlaton 6.023x103 E=hc

x = { 34 axies and = 6.625x10.flex 3x10 minit 25.9837 x 102*10*7. - 19.875 xoom 25.9837 = 0.7649 X10 = 764.9 om - 2.3 6.023x10 Q;, ca has E = 184.8 kJmol. latom, 184.8 = 30.6823x102 kJ E-hc ha = 6.625x10 / x X3x10 m. se 30.682341032x10 -34 6.615X10 se 19. 875 30.6823 6823 8106 = 0.64774106x109 mms X = 647.71m -23 (iiCase = 198.7 kJ/mol I atom, 198.7 = 32.990 VIDT - 6.023x1023 E=he → x=he -34 = 6.625x10 26x3x108 seat 32.990 x1023X103F. = 0.6024x106m X = 602.4 mm

) (a has E=191.4 kilmol latom → 191.4 -23 - 31.7781x10 - 6.023x10 3 E = har x= he = 6.625*108 ( X 3 X 1 set 31.778x107341031 = 0.6254X1060 E = 625.4 nm iv) la has E= 216.1 Ilmol tatom & 216.1 am Ende de 3 = 35.8791X1023 E = hc + s = ne -34 =61625x10 AtX3y106 miscel to' 35.8791X1521103y, = 0.55394xjö om = 553.9 om 1 n 6 Sy hae E = 178.210 Ilmol latom & 178.2 .. 61023x103 = 29.5865x1623 Eche x= he = 6.625x108 5xc X 3 X 1 set 21.5865x1023xios = 0.67175x106mg X=671.7 ml

34 1 ) Si hau E= 197.5 kJ/mol laton - 197.5 43 = 32.7909 x 10 23 k5 6,823 x 103 E = he > x = hc = 6.625x10 / SEE X 3 X 108 set - 3217909x 1023 103 0 = 0.6061 xroom. | X = 606.1*m7 6) Ba ;) has E = 144.5 kilnole latom = 144.5 = 23.991 X10 kJ Gibt 3x133 E = k + 1 = hệ = 6.625x1034 / x * 3x108 miscat 23.991x16731031 = 0.8284 x106m * = $26.40 -23 E (ii) E= 161.9 kil mole latom 161.9 2=26.880x10 kJ 23 1 6.023x103 E=hc > x= he = 6.625x103 4 X 3x10 museet 26.880x102310957 = 0.7393 xio bem -739-3 om

7. Element Energy (kg/mol) wavelength inm) 671.77 178-2 203.5 588.24 155.3 156.5 770.8 764.9 184.8 198.7 191.4 116. I 647.7 602.4 625.4 553-9 671.7 178.2 197.5 606.1 144.5 161.9 82814 739.3