Question

using bohrs atomic model derive expression fr calculating radius of orbit of helium

Answer 1

Derivation of Bohr’s Radius and Rydberg’s Constant The centripetal force between the electron and the nucleus is balanced by the coulombic attraction between them: 2 2 2 r Ze r mv = , where Z = atomic number and e = charge of electron Note that Coulomb’s constant is not required if the unit kg½m3/2s -1 is used for charge. Consequently, each expression will produce kgms-2, which is a Newton. Alternately, once can express velocity and mass using cgs(cm-g-s) and then use the esu for charge. 1esu = 3.33564 × 10−10 C 1 kg½m3/2s -1= 1.05482 X10 -5 C Since -1C = 6.241506×1018 electrons, then charge of e = -1.5189 9 X 10-14 kg0. 5 m1. 5 s -1 r Ze mv 2 2 = , (1) so r = 2 2 mv Ze (2) Since the electron’s energy is characteristic of its orbit, the energy cannot be lost or gained. It is quantized. According to Bohr, the angular momentum of the electron is a whole number n multiple of : 2π nh mvr = 2π nh , where v = velocity of electron; m = mass of electron; h = Planck’s constant v = mr nh 2π (3) Substituting (3) into (2): r mZe hn hmn rmZe mr nh m Ze r = = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = 22 22 22 2222 2 2 4 4 2 π π π (4) For the radius of hydrogen’s atom Z = 1 n = 1 h = 6.6262 X 10-34 Js = 6.6262 X 10-34 kg m2 /s2 (s) = 6.6262 X 10-34 kg m2 /s m = 9.1096 X 10-31 kg e = 1.5189 9 X 10-14 kg0. 5 m1. 5 s -1 r = ( ) = 2 31- 21-5 1.5 0. 14- 2 2 34- 1-2 kg)(1)(-1. 10 X 9.1096(4 ) s m kg10 X 4 5189 s m kg 10 X 6.62621 π 0.529 X10-10 m = 0.529 Ầ The total energy of the electron is the sum of its kinetic and potential energies: ET = Ek + Ep = mv 2 /2 - Ze2 /r (5) Substituting (1) into (5): ET = Ze2 /2r - Ze2 /r ET = Ze2 /2r - 2Ze2 /2r ET = -Ze2 /2r (6) Substituting (4) into (6): 22 422 22 22 2 2 4 4 2 hn emZ mZe hn Ze ET π π − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = = 2 2 422 2 hn emZ o − π When an electron falls back to a lower energy level, it emits a photon of energy hυ, which is the difference in the energy between the energy outer level, Eo and that of the inner level, Ei h υ = Eo - Ei = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − 2 2 422 2 2 422 2 2 hn emZ hn emZ o i π π = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − 2 22 422 2 11 nn oi h π emZ υ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − 3 22 422 2 11 nn oi h π emZ , where = − = − − − − 34 3 2 -31 2 415.15.014 3 422 106262.6( ) 2 1051894.1(kg)(1)10 X (9.10962 ) 12 smkgX smkgX h π emZ π 3.29 X 1015 s-1