Model of Hydrogen Atom

 K= 1/(4πε0) N-m2/coulomb2 * 109 Z (atomic no.) n (principal quantum no.) h (planck constant) joules-sec * 10-34 c (velocity of light) m/sec *108 G (universal constant of gravitation)nm2kg-2 or m3kg-1s-2 *10-11 e (charge of electron) coulomb *10-19 me (mass of electron) kg *10-31 (u can try by substituting me  with mπ- =273me& mn0 =16700*10-31 Find out Λ = Einstein Constant (8πG/c4) * 10-43 h/2π (reduced h) * 10-34 planck mass(mp):         √[hc/G] * 10-8 planck length(lp):        √[hG/c3] * 10-35 planck time(tp):           √[hG/c5] * 10-43 Gravitational Coupling constant of e-=Gme 2 /(hc)=(me/mp)2 * 10-46 LV=Kze2(angular momentum* velocity in a circular orbit) * 10-28 Rydberg Constant: R∞=K22π2 mee4/(ch3) m-1 * 105 u ratio=u1=(1/me)(h /e)2 u2=(1/4πK)*u1 * 10-10 Length of the string (L1=n2u2/z) * 10-10 Energy of a vibrating particle trapped in a box (vibration due to stretched string)E=n2h2/8ml2 Length :* 10-10  mE:            eV Fine Structure Constant: α=Ke2 /ch * 10-3       =(1/137) λe=h/(mec):compton wavelength of e- * 10-11     m fe=mec2 /h: compton frequency of e- * 1019     Hz ωe=2πfe:compton angular frequency of e- * 1019     Hz λde=h/(mev1):debroglie wavelength of e-for ground state n=1 * 10-11     m fde=v1 /λde: debroglie frequency of e- * 1019     Hz ωde=2πfde:debroglie angular frequency of e- * 1019     Hz v1:velocity in 1st Bohr orbit=2πKze2 /nh & n=1 * c  = αc/n where n=1 v2:velocity in 2nd Bohr orbit=2πKze2 /nh & n=2 * c  = αc/n where n=2 v3:velocity in 3rd Bohr orbit=2πKze2 /nh & n=3 * c  = αc/n where n=3 v4:velocity in 4th Bohr orbit=2πKze2 /nh & n=4 * c  = αc/n where n=4 Bohr Magneton: eh/(4πme) Am2 *10-24 r1:radius of 1st Bohr orbit=n2h2/m4π2Kze2  & n=1 *10-10  m     (0.53Angstrom for hydrogen) r2:radius of 2nd Bohr orbit=n2h2/m4π2Kze2  & n=2 *10-10 r3:radius of 3rd Bohr orbit=n2h2/m4π2Kze2  & n=3 *10-10 r4:radius of 4th Bohr orbit=n2h2/m4π2Kze2  & n=4 *10-10 r21=r2-r1=3r1r32=r3-r2=5r1 r43=r4-r3=7r1 r54=r5-r4=9r1 *10-10 *10-10 *10-10 *10-10 total energy:-2mπ2K2z2e4 /n2h2 E1 at n=1 (ground state) *10-18J = eV E2 at n=2 (first excited state) *10-18J = eV E3 at n=3 (second excited state) *10-18J = eV E4 at n=4 (third excited state) *10-18J = eV E∞  at  n=∞ zero Change in total energy = 2mπ2K2z2e4 /h2   (1/n12   - 1/n22 ) = hc/λ or 1/λ =R(1/n12   - 1/n22 ) Lyman series, λα of first member,n1=1 & n2=2 *10-10m =4/3R Balmer series, λα of first member,n1=2 & n2=3 *10-10m =36/5R Paschen series, λα of first member,n1=3 & n2=4 *10-10m =144/7R Bracket series, λα of first member,n1=4 & n2=5 *10-10m =400/9R Pfund series, λα of first member,n1=5 & n2=6 *10-10m =900/11R Humprey series, λα of first member,n1=6 & n2=7 *10-10m =1764/13R Ke2 / mc2 *10-15m (diameter of nucleus) h2 /Kme2 *10-10m (diameter of atom) T1:time period in 1st Bohr orbit=2πrn/vn & n=1 *10-18s T2:time period in 1st Bohr orbit=2πrn/vn & n=2 *10-18s T3:time period in 1st Bohr orbit=2πrn/vn & n=3 *10-18s T4:time period in 1st Bohr orbit=2πrn/vn & n=4 *10-18s Mean life *10-8s No. of Revolution before decay (in T1) *1010 No. of Revolution before decay (in T2) *1010 No. of Revolution before decay (in T3) *1010 No. of Revolution before decay (in T4) *1010

 * To prove  L=angular momentum=mvr=nh/2π   mv2/r=kze2/r2  or (mvr)v=kze2  or Lv=kze2Ek =mv2 /2  =kze2 /2r   Ep =-kze2 /r =-mv2TE =-kze2 /2r =-mv2 /2 =-mk2z2e4   /2L2 . Differentiating, d(TE) =(mk2z2e4   /L3 )dL=(mv2L2 /L3)dL=mv2dL/L =hν=h/T=hv/2πr  or mvr*dL/L  = h/2π or dL=h/2π or L=nh/2π * To find out value of r, v, TE,λ Lv=kze2  or (nh/2π)v= kze2  or v=2πkze2 /nh = z*( 2πkze2/h)*(1/n)=z*α /n r= kze2 / mv2 or r =n2h2  / 4π2mkze2TE=-mv2 /2 =-2π2mk2z2e4   /n2h2 ΔTE=hc/ λ =(2π2mk2z2e4/h2) [1/n12  - 1/n22] or 1/λ =(2π2mk2z2e4/ch3) [1/n12  - 1/n22]=(2π2mk2e4/ch3) [1/n12  - 1/n22] when z=1 or 1/λ =R [1/n12  - 1/n22]  where R=(2π2mk2z2e4/ch3) * To find dimension of the atom From de broglie hypothesis, λ = h/p = h/mv We know that when a string of length l  is tied to a rigid support at both ends and is stretched, it vibrates in such a manner so that the nodes are at both ends and the condition that is satisfied is sin (2πl/λ) =0 or  2πl/λ =nπ where n is any integer for standing wave to be formed.  λ = 2l/n or l=n (λ/2) Hence, 2l/n=h/mv or p= nh/2l Taking l=2πr, we get 4πr =nh/mv or mvr=nh/4π E=p2/2m = n2h2/8ml2 .....(a). This is the total energy of a vibrating particle trapped in a box of length l. But E=2π2mk2z2e4   /n2h2   ....(b)   ; So equating (a) & (b), we get   l=n2h2  /4πmkze2 whose dimension is of the order of 10-10 m.If in the Bohr equation, we substitute mass of electron by mass of neutron, the dimension comes to the order of 10-14m   which is roughly the size of nucleus. * To prove mvr=nh/2π from de broglie wavelength We already know that the condition for standing wave in case of a stretched string of length l tied rigidly to both ends is l=n (λ/2) de Broglie considered the circumference of Bohr's circular orbit as a string of length l=2πr where standing waves are formed. He assumed that these are not ordinary waves but matter waves of energy E=pc where p is the momentum & E = hc/λ . This follows from Einstein's Special theory of relativity E=√(m02c4 +p2c2). For entities with zero rest mass E=pc Equating E from both cases, λ=h/p Here de broglie enters the scenrio , where he postulated the symmetry between mass less entities and entities having non-zero rest mass and extended the concept to λ=h/p where p=mv for these entities. Hence , this was a postulate and conclusion was not reached through derivation. Now, l= nh/(2mv) or 2πr= nh/(2mv) or mvr=nh / 4π. or angular momentum = h/4π, 2h/4π,3h/4π, ....... de Broglie assumed l=n λ . Why ??? possible explanation is that when matter is converted to energy, there is equal contribution of matter and anti-matter. Hence anti-matter also contributes anti-matter angular momentum=  h/4π, 2h/4π,3h/4π, ....... Hence total angular momentum of matter wave=  h/2π, 2h/2π,3h/2π, .......nh/2π or mvr=nh/2π (this is not in any text book but my assumption) * de broglie wavelength is velocity dependant whereas compton wavelength is not velocity dependant.