Model of Hydrogen Atom

  Boltzman Constant = KB = m2kg/s2k * 10-23  
  K= 1/(4πε0) N-m2/coulomb2 * 109  
  ε0=1/4πK * 10-10  
  magnetic permeability of vacuum μ0 * 10-6  
  ε0μ0 * 10-16  
  ε00 * 10-4  
  Z (atomic no.)  
  Z' (no. of times electron mass)  
  n (principal quantum no.)  
  h (planck constant) joules-sec * 10-34  
  h=h/2π * 10-34  
  c (velocity of light) m/sec *108  
  ch (unit energy-length) * 10-26  
  h/c= * 10-42   (dimension ML)  
  h/c= * 10-42   (dimension ML)  
  G (universal constant of gravitation)nm2kg-2 or m3kg-1s-2 *10-11  
  L/M=G/c2  ; (dimensional analysis) * 10-27  
  L / (Q2/M)=K/c2  ; (dimensional analysis) * 10-7  
  GK/c4 * 10-34  
  √2(GK/c4 ) * 10-34  
  K/G *1020  
  c/μ0 *1014  
  c/K *10-1  
  c/ε0 *1018  
  electron Bohr radius (classical) er = *10-15  
  electron mass density(classical) mr = *1012  
  proton charge radius    pr= *10-15  
  e (charge of electron) coulomb *10-19  
  me (mass of electron) kg *10-31  
  Electron Rest energy =me c2 *10-16  
  h /me=v1r1 *10-4  
  e*√(Kme) *10-30  
  (1/2)Kq2 = *10-29  
  hc/me  ; *105  
  e/me =2*gyro magnetic ratio for L=2*(e/2me ) *1012  
  (e/me)2 = *1020  
  dgm1(dielectric gyro magnetic constant)=K*e/me *1021  
  mP/me *103  
  me/mP *10-3  
  cdgm0=Lv=KZe2 *10-28  
  cdgm0r=√(cdgm0)=√(KZe2) *10-14  
  cdgm1=KZe2 /me =K*Ze*(e/me)=dgm1*Ze *102  
  √(KZe2 /me)=cdgm3=(2TE/L)*r3/2 *101  
  cdgm1/3π (roughly equal to 33) *102  
  At what velocity, the relativistic mass of e will be equal to proton rest mass c  
  bijan=(K/G)*(e/me)2 =(this is a dimensionless quantity) *1042  
  Magnetic dipole moment in orbit 1, M1a=-(L/2m)*e=--(e/2me)*L *10-22  
  mass of the proton  mP *10-27  
  me*mP *10-58  
  mGP=√(me*mP) *10-29  
  (u can try by substituting me  with mπ- =273me& mn0 =16700*10-31    
  1Joule = ? eV * 1018  
  Find out    
  Josephson constant Kj = 2e/h   * 1015  
  Josephson constant Kj =√[8α /(μ0*c*h)]   * 1015  
  Entropy per unit area=KBc3/4Gh   * 1046  
  Λ = Einstein Constant (8πG/c4)   * 10-43  
  h=h/2π (h=reduced h) * 10-34  
  electron classical volume * 10-44  
  electron spin density=σ =h/(2*electron volume) * 1010  
  planck mass(mp):         √[hc/G] * 10-8  
  bijan mass(mb):         √[hc/K] * 10-17  
  planck energy(mpc2) : * 108  
  planck length(lp):        √[hG/c3] * 10-35  
  bijan length(lb):        √[hK/c3] * 10-24  
  planck length/planck mass= * 10-27  
  bijan length/bijan mass= * 10-7  
  (planck length/planck mass)*(bijan length/bijan mass)*√2 * 10-34  
  planck time(tp):           √[hG/c5] * 10-43  
  planck density(dp)      c5 / hG2 * 1028  
  planck temperature in K =(mpc2)/KB * 1032  
  Planck charge=√(hc/K) * 10-18  
  Gravitational Coupling constant of e-

=Gme 2 /(hc)=(me/mp)2

* 10-46  

(angular momentum* velocity in a circular orbit)

* 10-28  
  h/α=L(V/c)=(K/c)ze2 * 10-36  
  Virtual velocity=Vvir =k e2/h * 108  
  Virtual Kinetic Energy=Econ=(1/2)meV2vir * 10-18  
  Rydberg Constant: R=K22 mee4/(ch3) m-1 * 105  
  Energy constant=Econ=K22 mee4/(h2)=-E1


* 10-18  
   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)


Length :* 10-10  m

E:            eV

  Fine Structure Constant: α=Ke2 /ch * 10-3       =(1/137)  
  α5 * 10-15   
  electron orbit radius =re2=  α5*(hc/me ) * 10-10   
  mec (dimension MLT-1) * 10-23   
  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  
  classical electron orbit radius r0=Ke2/( me *c2) *10-15  
  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 =4r1, & n=2 *10-10  
  r3:radius of 3rd Bohr orbit=n2h2/m4π2Kze2 =9r1, & n=3 *10-10  
  r4:radius of 4th Bohr orbit=n2h2/m4π2Kze2 =16r1, & n=4 *10-10  








  total energy:-2mπ2K2z2e4 /n2h2    
  E1 at n=1 (ground state)=-2mπ2K2z2e4 /h2 *10-18J = eV  
  E2 at n=2 (first excited state)=E1/4 *10-18J = eV  
  E3 at n=3 (second excited state)=E1/9 *10-18J = eV  
  E4 at n=4 (third excited state)=E1/16 *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  
  KE1=-E1 *10-18J = eV  
  KE2=-E2 *10-18J = eV  
  KE3=-E3 *10-18J = eV  
  KE4=-E4 *10-18J = eV  
  PE1=2E1 *10-18J = eV  
  PE2=2E2 *10-18J = eV  
  PE3=2E3 *10-18J = eV  
  PE4=2E4 *10-18J = eV  
  presumptive acceleration a1 at 1st Bohr Orbit (a1=v12/r1) actual acceleration=0 since formula a=v2/r is invalid for a *1022  
  presumptive acceleration a2 at 2nd Bohr Orbit (a2=v22/r2)  actual acceleration=0 since formula a=v2/r is invalid for a *1022  
  presumptiveacceleration a3 at 3rd Bohr Orbit (a3=v32/r3)  actual acceleration=0 since formula a=v2/r is invalid for a *1022  
  presumptive acceleration a4 at 4th Bohr Orbit (a4=v42/r4) actual acceleration=0 since formula a=v2/r is invalid for a *1022  

As per classical electrodynamics, accelerated charged particles always radiate electro magnetic waves. . Since the electron is accelerating, a classical analysis suggests that it will continuously radiate energy, and therefore the radius of the orbit would shrink with time and ultimately it shall fall into the nucleus. In Reality, It does not radiate energy due to quantum Mechanical considerations. Below we have calculated presumptive power radiation in Gaussian units as per Larmor formula below.

  Average radius a0 at classical Bohr Orbit


  Presumptive time t1 to fall into the nucleus

 t1=a03 /(4r02c)

  Power radiated 1st orbit P1=2e2a12/(3c3) *10-18  
  Power radiated 2nd orbit P2=2e2a22/(3c3) *10-18  
  Power radiated 3rd orbit P3=2e2a32/(3c3) *10-18  
  Power radiated 4th orbit P4=2e2a42/(3c3) *10-18  
  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 2nd Bohr orbit=2πrn/vn & n=2 *10-18s  
  T3:time period in 3rd Bohr orbit=2πrn/vn & n=3 *10-18s  
  T4:time period in 4th  Bohr orbit=2πrn/vn & n=4 *10-18s  
  ν1:frequency in 1st Bohr orbit=2πrn/vn & n=1 *1015s  
  ν2:frequency in 2nd Bohr orbit=2πrn/vn & n=2 *1015s  
  ν3:frequency in 3rd Bohr orbit=2πrn/vn & n=3 *1015s  
  ν4:frequency in 4th  Bohr orbit=2πrn/vn & n=4 *1015s  
  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  
  Orbital Angular momentum in 1st Bohr Orbit L1= h * 10-34  
  OrbitalAngular momentum in 2nd Bohr Orbit L2=*2 * 10-34  
  Orbital Angular momentumin3rd Bohr Orbit L3=3h *3 * 10-34  
  OrbitalAngular momentum in 4th Bohr Orbit L4=4h *4 * 10-34  
  spin angular momentum of e in Bohr orbit=|S|=        √[s(s+1)]*h=√3/2  * h * 10-34  
  z component of Spin Angular Momentum=Sz =s/2 = h/2 * 10-34  
  Gyro magnetic ratio for spin =(e/m) *1012  
  current in 1st Bohr Orbit=I1=e/T1=eν1 *10-3  
  current in 2nd Bohr Orbit=I2=e/T1=eν2 *10-3  
  current in 3rd Bohr Orbit=I3=e/T1=eν3 *10-3  
  current in 4th Bohr Orbit=I4=e/T1=eν4 *10-3  
  Area of 1st Bohr orbit  A1=πr12 *10-20  
  Area of 2nd Bohr orbit A2=16r12=πr22 *10-20  
  Area of 3rd Bohr orbit A3=81r12=πr32 *10-20  
  Area of 4th Bohr orbit A4=256r12=πr42 *10-20  
  Magnetic dipole moment, orbit 1, M1=I1A1 *10-22  
  Magnetic dipole moment, orbit 2, M2=I2A2 *10-22  
  Magnetic dipole moment, orbit 3, M3=I3A3 *10-22  
  Magnetic dipole moment, orbit 4, M4=I4A4 *10-22  
  E1/L1 *1016  
  v1/r1 (-2E/L =v/r =ΔE/ΔL for all orbits derived classically) *1016  
  L12 /E1 *10-50  
  -2me r12 ; (for all orbits L2 /E=-2me r2  or L2 =-2meE*r2) *10-50  
  -mK2z2e4 /2=E1L12 *10-86  
  -mK2z2e4 /2=E2L22 *10-86  
  Linear Momentum in 1st Bohr Orbit=P1=L1/r1 *10-24  
  TE1=P12 /(2*me) *10-17  
  β1=v1/c *10-2  
  β2=v2/c *10-2  
  β3=v3/c *10-2  
  β4=v4/c *10-2  

Macroscopic Bar magnet is nothing but all the atomic magnets predominantly aligned in the same direction. Elementary atomic magnets are tiny circulating currents. Quantum mechanically , it is not only the orbital motion but also the spin of the electrons that contribute to atomic magnetism. The ratio of magnetic dipole moment M and orbital angular momentum M/L is e/2me( which is called gyro magnetic ratio) and the same is derivable classically as well as quantum mechanically. M =IA=(e/T)*πr2 . L=mvr=m *(2π r/T)*r=m2π r2/T  .So M/L=e/2m. But for spin, M/L is (e/m) in Quantum mechanics which is correct and e/2m in classical mechanics which is wrong. M and L are both perpendicular to the orbit and they are anti parallel to each other.

  Gravitational Force between proton & electron in 1st Bohr Orbit *10-47  
  velocity in 1st Bohr orbit due to gravitational effect=(GmP/r1)  *10-22 c       ---*10-14  
  Orbital angular momentum due to gravity in 1st Bohr orbit:me*v1*r1 *10-55  
  Time period due to gravity in 1st Bohr orbit *104  
  Frequency due to gravity in 1st Bohr orbit *10-4  
  total energy E1G at n=1 (ground state) due to gravity=-GMm/2r1 *10-59J = *10-40eV  
  E1G/(h/c)=Bi1 (dimension LT-2) *10-17  


* To prove  L=angular momentum=mvr=nh/2π

   mv2/r=kze2/r2  or (mvr)v=kze2  or Lv=kze2

Ek =mv2 /2  =kze2 /2r  

Ep =-kze2 /r =-mv2

TE =-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

The essential part of derivation was d(TE)/dL =2πν=2π / T = ω = angular velocity  and TE/L =-(1/2) v/r =-ω /2

We can also put 2TE*L2  =constant, differentiating and equating to zero, we get

ΔTE*L2 +2L ΔL *TE=0   or  ΔTE / ΔL =-2TE/L=v/r =(kze2/m)*r - 3/2

 Mass, charge being known, out of (L , TE, v, r ) you fix the value of any one, and can get value of other 3 by using 2 blue equations)

We find that quantum mechanically TE/L=v/r and not v/2r,

TE=hν =h/T =hv/2π r=h/2π*(v/r)=h*(v/r)=L*(v/r)

or TE/L =v/r

v2r =kQq/me  or ν2r3 =kQq/4π2me  or ν2V =(1/3)kQq/πme where V is the volume of sphere  with radius r and ν is frequency of revolution. These are classically derived results. Taking Q=ze=1*e,   ν2V =(1/3π)*K*e*(e/me) =26.83

We also get L=√(mKZ) √r *e

Another way to look at the above problem :

In gravitation as well as in Bohr orbits, the treatment is similar and classical. In gravitational orbits, we know the radius r and hence calculate the total energy and angular momentum which are constants since there is conservation law for both. But in Bohr orbit , we go the other way round and first fix the angular momentum and then calculate speed from LV formula and radius  etc and finally total energy. We invoke the Heisenberg uncertainty principle Δp Δx =nh  or L=  pr =nh  and then proceed. By making the orbital velocity constant via radius which is also a constant , we eliminate the problem of accelerating charge and consequent energy loss due to radiation. To elaborate , in uniform circular motion, speed is constant but not the direction of velocity. This change of direction produces acceleration  which is v2/r. A charge when accelerated looses energy by radiation which results in reduction of radius till the charge falls on the nucleus. To eliminate this, Bohr proposed zero acceleration which means the velocity both magnitude and direction to remain same throughout, which implies Δr =0

 Bijan's Derivation:

In the above derivation, we have equated centripetal force with electrostatic force. Actually, the force exerted on a moving electron is not the coulomb force and moreover electro dynamic force is not a conservative force where as Coulomb force is conservative. Immediate consequence of this is non conservation of angular momentum which is contrary to our assumption. Moreover, both charge and mass are mixed up . We adopt a pure gravitational approach here.

In gravitation, when the planet revolves in a closed circular orbit (which is a special case of elliptical orbits and the same is taken as per Bohr assumption), the force is conservative, KE is equal to the TE with sign changed, and Potential energy is numerically 2 times the kinetic energy .

So KE= constant

PE=-2KE =-mv2 =-mv*v=mv*(2πr/T)=-(mvr)*(2π/T) =-(mvr)*(2πν)

Since PE is constant times the TE in a circular orbit, conservation of TE implies conservation of PE.

So PE=const and (mvr)*(2πν) =const .

mvr=(const/2π)*(1/ν)=C*(1/ν). The constant ,say C should have the dimension of energy or negative energy if we define -sign suitably.

If const=C=(nhν/2π)=E/2π,


Here conservation of Angular momentum follows from conservation of potential energy PE=-2nhν=constant and conservation of TE=- nhν

In a closed circular orbit, there are 4 parameters  TE, L , r , v . At every point in the orbit, TE and L are conserved. Also r, v are conserved in magnitude but not in direction, However rv is conserved both in magnitude as well as in direction which implies that r, v are in a fixed plane. Out of the 4 parameters, we can assign (classically) any arbitrary value to any one of the parameters and the other 3 can be expressed in terms of this parameter.

case 1: Assign any value to L

v=kze2/L       r=L2/kze2m      E=-(kze2)2 m  /(2L2)

acceleration=a=v2/r=(kze2)3m /L4 ;Putting L=h , we get a=v2/r=(kze2)3m /h4  ;

case 2: Assign any value to E

L=√m*kze2/√(-2E)       v= √(-2E) /√m     r=(kze2) /(-2E)

acceleration=a=v2/r=4E2 /kze2m ; Putting -2E=(kze2)2m /L2 = (kze2)2m /h2 ; a=(kze2)3m /h4  ;

case 3: Assign any value to v

L=kze2/v    r = (kze2) /(mv2)    E=- mv2/2

acceleration=a=v2/r=mv4 /kze2; Putting v=(kze2) /L = (kze2)/h ; a=(kze2)3m /h4  ;

case 4: Assign any value to r

v= √(kze2) /√(mr)    L=√(kze2mr)     E=kze2/2r

acceleration=a=v2/r=kze2 /mr2; Putting r=L2/(kze2)m  = (kze2)/h ; a=(kze2)3m /h4  ;

Since circle is a special case of ellipse, In elliptical orbits which are generalized cases, case 3and 4 are not applicable since v,r change from point to point. However, in elliptical orbits, conservation of E does not automatically ensure conservation of L or vice versa. A third component is to conserved for conservation of both E and L If we start from E, conservation of eccentricity e is essential for conservation of L. If we start from L, we observe that a , e need not be individually conserved for L bur a*√(1-e2) =b should be conserved. But for conservation of E, a only should be conserved. Hence for both, a and e should be individually conserved.

* 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  / 4π2mkze2

TE=-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)

Comparison between Circular Orbit in Gravitation & Bohr Orbit :

Gravitation Circular Orbit                                                                                          Bohr Orbit

(1) TE=- GMm/2r                                                                                                       TE=-KQq/2r

(2)PE= - GMm/r                                                                                                          PE=-KQq/r

(3)KE=  GMm/2r                                                                                                         KE=KQq/2r

(4)L =m (GM/r) r                                                                                                      L =m (KQq/mr) r

        =m m (GM) r                                                                                                  =m q (KQ) r

(5)P =  m (GM/r)                                                                                                       P =  m (KQq/mr)

         = m m (GM/r)                                                                                                  = m q (KQ/r)

(6) (1/2)mv2 gives unit of KE                                                                                     (1/2)qv2 gives unit of KE

(7)Dimensionally G=L3T-2/M  = L2T-2*(L/M)                                                        K=ML3T-2  / Q2 ={L/(Q2/M)} L2T-2

     Or G/c2 =L/M                                                                                                        K/c2 =L/(Q2/M)

     LHS of the order of 10-27                                                                                     LHS of the order of  10-7 ;

(G/c2) *(K/c2) is of the order of 10-34  which is same as h and if we further multiply by √2, becomes almost equal to h up to 2 decimal points.

(8)TE * L2 =-(1/2)mG2Z12m4                                                                                  TE * L2 =-(1/2)mK2Z2e4   (Z,Z1 are integers)


We observe that the energy formulation is similar for gravitation and bohr orbit.  But there is a difference in the formulation of angular and linear momentum marked by red. In bohr orbit, one mass is substituted by charge.
  gravity electrostatics value gravity value electrostatics gravity*electrostatics gravity*electrostatics
L2 [hG/c3] [hK/c3] 2.6122963*10-70 0.0352259*10-48 [h2GK/c6] 0.0920204*10-118
L √[hG/c3] √[hK/c3] 1.61626*10-35 0.1876859*10-24 √[h2GK/c6] 0.3033486*10-59
M2 [hc/G] [hc/K] 4.736942*10-16 0.0351285*10-34 (h2c2/GK) 0.1664016*10-50
M √[hc/G] √[hc/K] 2.1764517*10-8 0.187426*10-17 (h2c2/GK) 0.4079235*10-25
L2/M2 G2 /c4 K2 /c4 0.551473*10-54 1.0027751*10-14 G2K2/c8 0.5530018*10-68
L/M G /c2 K /c2 0.7426123*10-27 1.0013866*10-7 GK/c4 0.7436409*10-34
(L/M)gravity /


G/K L is length, M is mass 0.0741584*10-19      
Mem/Mgravity     0.0861153*10-9      
(L/M3)combined  G2 K2 /h2c6   4.4689526*1016      

* 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.

*Electric Field outside a current carrying Conductor :

Why an electric field is not produced  outside a current carrying conductor ? It depends upon what type of current is flowing through the conductor. Suppose a constant current is flowing. Take any cross section of the wire.

the amount of charge leaving the cross section=the amount of charge entering the cross section. So the whole wire is electrically neutral and no electric field is produced. But consider a circuit with AC source.

that variable current will produce variable magnetic field. If you will examine it carefully, electric field and magnetic field will be produced perpendicularly to each other and will be variable of course. So, now this will become the source of Electromagnetic Wave. Because when changing electric field and changing magnetic field is produced at a point simultaneously and perpendicular to each other, the electromagnetic wave is generated which will move in the direction perpendicular to Electric and magnetic field. To be specific, the direction of wave is given by the cross product of E and B.