Bohr's theory of Hydrogen atom-7

 Redoing Bohr's Theory of Hydrogen atom

       When the electron is far away from the proton ,it has no kinetic and potential energies but has its own mass energy moc2 . When it approches the proton due to attractive nuclear force  it gets accelerated towards the nucleus and gains  kinetic energy and mass due to continuous changes in its velocity . The amount of workdone by the electron is given by                                                                                                                                                                                                               

                                    W =  ∞ ∫rn  F.dr  =  ∞ ∫rn  [e62/Kr^2]dr = - e^2/Krn 

As it happens in the conservative force field, the loss of potential energy  of the electron is converted into its kinetic energy and relativistic increase of mass   e^2/Krn =  (1/2)mnvn^2 + dmc^2  or e^2/2Krn  = dmc^2. It shows that half of the loss of potential energy is converted into kinetic energy and the remaining half is deposited as added mass to the electron. In the micro-vacuum space surrounding the nucleus  the electron sustains its  motion without loosing its kinetic energy .   On the way of finding stability, the electron undergoes a circular motion, where the nuclear attractive force is counter-balanced with centrifugal force i.e.,e^2/Krn^2  =  mnvn^2/rn,  

      When the electron's speed increases, its relativistic mass increases, because the energy used to accelerate it is added to its total energy, which is then converted into added mass. This effect is negligible at low speeds but becomes significant near the speed of light.  Let the mass of the electron  be increased from mo to mn due to its relativistic velocity.

dmc^2 = (mn - mo)c^2 = [mo(1+ vn^2/2c^2) - mo]c^2  = (1/2)movn^2 = (1/2)mnvn^2 (mo/mn)

          = [e^2/2Krn](1-vn^2/c^2)^1/2  = (1/n^2)[e^2/2Kao] 

 When the electron with added mass is fixed in an orbit it is bound with the nucleus, the atomic  binding energy comes from the mass defect. In electromagnetic field it is contributed by the moving electron. Atomic binding energy of the hydrogen atom is the minimum energy required to completely disassemble into the  proton and the electron, the two constituents of the atom. It acts as a kind of "glue" holding the orbital electron and the proton (nucleus) together. This energy arises from the mass defect, where the total mass of constituent particles is higher than the mass of the atom formed by them, with the difference converted into energy (E=mc^2). This energy is radiated out as electromagnetic radiation. Thus the electron is bound to the nucleus by electromagnetic forces, with  a binding energy which is negligible compared to the MeV scale of nuclear binding energies. The energy required to bind the orbital electron with the nucleus comes from the interaction between the orbital electron and the nucleus. Atomic binding energy of hydrogen = [mass of proton + mass of orbital electron] -mass of hydrogen atom which is equal to sum of masses of free proton and free electron.

                        The mass of free proton mp = 1.67262192 x 10^-27 kg                                                                                   The mass of  free electron meo =  9.1093837 x 10^-31 kg                                                         Total mass of unbound hydrogen atom= mp  + meo = 1.67353286 x 10^-27 kg                              The mass of the orbital electron men = meo/(1- vn^2/c^2)^1/2  ≃ meo  + (1/2)meo vn^2/c^2                                                         The excess mass  =  (1/2)meo vn^2/c^2  

When the orbital electron is bound with the nucleus its binding energy comes from its mass

          Binding energy  = (1/2)meo vn^2  = (1/2)men vn^2 x meo/men = [e^2/2 n^2Krn](1-vn^2 /c^2 )^1/2 

                                                               =(1/n^2) [e^2/2 Kao]

This is the sum of  the potential and kinetic energies of the orbital electron . This energy is radiated out from the atom.

                                                    (1/n^2) [e^2/2 Kao]  = hv λ∞→n                                                                The kinetic energy of the electron = (1/2) mn vn^2, where mn is the mass of the electron moving with velocity vn. 

Potential energy of the electron = - e^2 /Krn

mass equivalent energy = mnc^2

Total energy content of the electron at n1, E(n1) = (1/2) mn1 vn1^2 - e^2 /Krn1 + mn1c^2 = - (1/2) e^2 /Krn1 + mn1c^2

Total energy of the electron at n2,  E(n2) = (1/2) mn2 vn2^2 - e^2 /Krn2 + mn2c^2 = - (1/2) e^2 /Krn2 + mn2c^2

When the electron jumps from n2 to n1  the transition energy is E(n2) - E(n1)

                [- (1/2) e^2/Krn2 + mn2c^2] - [ - (1/2) e^2/Krn1 + mn1c^2]  = hν2→1 

       (1/2) e^2/K [ 1/rn1 - 1/rn2 ]  - [mn1 - mn2]c^2 =    hν2→1  

 Considering the innermost orbit  in the hydrogen atom and the free electron               

  (1/2) e^2 /Kr1  - [ m1 - mo] c^2 =  hν∞→1

  (1/2) e^2 /Kr1 - (1/2) mo v1^2 =   (1/2) e^2 /Kr1 - (1/2) m1 v1^2  (mo/m1) =   (1/2) e^2 /Kr1 [ 1- 1 + v1^2/2c^2 ]

                                   =   (1/2) [e^2 /Kr1] ( v1^2 /2c^2 )

    e^2 /Krn  - e^2/2Krn =  e^2 /2Krn =[mn -mo]c^2  = c^2 [mo  - mo (1-vn^2 /c^2 )^1/2] /(1-vn2 /c2 )1/2                                                                                                                                                          = (1/2) movn^2  /(1-vn^2 /c^2 )^1/2  = (1/2) mn vn^2 = (1/2) e^2 /Krn

But dm c^2 = (1/2) mo vn^2  = (1/2) mn vn^2 (mo/mn) =(1/2) [e^2 /Krn](1 - vn^2/c^2 )^1/2 = e^2/2Kao 

The decrease in potential energy is converted into kinetic and mass energy of the orbital electron

Tamil poem

 இந்த உலகம் தாங்காதைய்யா தாங்காது

இன்னொரு போரைத் தாங்காது 

முதலிரு போர்கள் விட்டுச் சென்ற சுவடுகள்

இன்னும் நெஞ்சைவிட்டு அகலவில்லை  

இன்றும் தொடரும்  விதிமீறிய  இன்னல்கள் 

இறைவனே இன்னும் அழுது ஓயவில்லை

பெற்ற பிள்ளைகளை மீண்டும் இழக்கலாமா-இறுதிவரை  

உற்ற துணைகள் உதவிக்கு இல்லாமற்போகலாமா 

கட்டிய வீடு கண்முன்னே இடிந்து போகலாமா

ஒட்டிய பொருட்கள் எரிந்து சாம்பலாகலாமா

பசியோடு பலநாட்கள் பருகத் தண்ணீரில்லை 

நாடுஇருந்தும் நாடோடியாய் நடைப்பயணம் 

அங்கே அறிவுக்கும்   பொருளுக்கும் மதிப்பில்லை 

மக்களை ஆளவந்த அறிவிலிகளே, 

புத்தியை அடகு வைத்துவிட்டா நாடாளாவந்தீர்கள்

இந்த உலகம் தாங்காதைய்யா தாங்காது

இன்னொரு போரைத் தாங்காது 

 Revisiting Bohr's Theory of Hydrogen Atom

    The hydrogen atom with a central proton and an orbiting electron is a two body system. When a  free electron enters the active space it gets accelerated towards the proton , thereby its kinetic energy is increased  gradually due to accelerated motion towards the nucleus and simultaneously it gains some mass due to relativistic variation  of its velocity .The energy required for this comes from the source that accelerates the electron. The work done by the electron reduces its potential energy. In this motion of the electron within the atom, the energy must be conserved at every point of its path. It means that the loss of its potential energy  must be equal to gain in its kinetic energy and the energy used for the relativistic increase of mass. That is a fraction of the loss of potential energy is converted into its kinetic energy and remaining is stored as its relativistic increase of mass, so that the energy is conserved all along it path. As the accelerated electron is abruptly stopped at any one of the allowed orbits, it takes up a curved path until it attains stability in a stable orbit. It revolves round the proton in circular orbit, where the electrostatic force of attraction is exactly counter-balanced with its centrifugal force The conservation of energy of the electron at all of its position  requires that the loss of its potential energy  must be equal to gain in its kinetic energy and the energy used for the relativistic increase of mass.

                            e^2/Krn  =  (1/2) mn vn^2   +  Δm c^2    where  Δm = (mn - mo)

  The dependency between the orbital velocity, radius and the quantum number is little changed due to relativistic variation of  mass of the orbital electron.

     When the electron is attracted by the nucleus it gets accelerated. Since the force is inversely proportional to intermediate distance, the acceleration experienced by the electron is not uniform, but variable. Let mn, rn , vn  be the mass, radius and velocity of the electron in its nth orbit, and an is the acceleration experienced by the electron when it jumps towards the nucleus. 

F =  mn an = e^2/Krn^2 = mn vn^2 /rn  or an = vn^2 /rn

It shows that  e^2/K = mn an rn^2  = mn vn^2 rn = [mo/(1- vn^2 /c^2)^1/2] vn^2 /rn . By substituting the value of rn = n^2ao(1- vn^2 /c^2)^1/2  we get  e^2/K =  mo vn^2 n^2 ao  = constant or vn n = constant irrespective of the relativistic variation of mass of electron.

The variable acceleration an  = vn^2 /rn = [vn^2/(1- vn^2 /c^2)^1/2][1/n^2 ao].It is worth to find out how far the dependency of variables in Bohr's theory of hydrogen atom changes due to relativistic increase of mass of the orbital electron.  

(1).Orbital velocity (vn) Vs Orbital radius (rn)

      Let us suppose that an electron in the hydrogen atom is in its nth orbit having radius rn. The nuclear attractive force is counter-balanced with the centrifugal force.

                                e^2/Krn^2  = mnvn^2/rn

                        or    e^2/moK   =  vn^2 rn/(1- vn^2 /c^2)^1/2 =  vn2 n^2 ao  = constant                             

where vn is the velocity of the electron in the orbit and mn is its relativistic mass. The relativistic variation of mass of the orbital electron makes no correction in the dependency of vn on rn.It implies that the added mass of the electron in the orbital motion is converted into atomic binding energy which holds the orbital electron with the nucleus.

(2).Orbital velocity ( vn) Vs Quantum number (n)

       According to Bohr's postulate, the angular momentum associated with the  orbital electron must be an integral multiple of h/2π 

                                   mn vn rn  = nh/2π                                                                                     

where n is quantum number characterizing the orbit.

Dividing one by the other,    vn  =  2π e^2/nKh  = e^2 /2nεoh                                                        

That is the dependency of vn  on n is not altered  even with the relativistic variation of mass of the orbital electron.

(3) Orbital radius (rn) Vs Quantum number (n)

By squaring      mn^2vn^2 rn^2  = n^2h^2/4π^2    

Dividing  by mnvn^2 rn ,  mn rn  = n^2h^2/4π^2x(K/e^2) 

[mo/(1-vn^2 /c^2)^1/2 ] rn =  n^2h^2 εo /πe^2 = n^2 moao

rn =   [n^2h^2 εo/moπe^2] (1-vn^2 /c^2)^1/2) = n^2 ao (1-vn^2 /c^2)^1/2 

Very same result can be obtained from the discreteness of the electronic orbits in atom. 

  2πrn  = nh/mnvn

squaring both sides and substituting the value for mnvn,  4π^2 rn^2   = n^2h^2/mn^2vn^2 = n^2 h^2 K rn/mne^2

or  rn =  n^2 ao (1-vn^2 /c^2)^1/2 

The relativistic variation of mass of the orbital electron makes the orbit to shrink about its center. Approximately the change in the radius is h^2/8π^2 m0^2 ao c^2 irrespective of n.

Substituting this value of rn in  mn vn rn  = nh/2π

[mo/(1-vn^2 /c^2)^1/2 )  x vn x n^2 ao (1-vn^2 /c^2)^1/2   = nh/2π

                      vn  =  h/n 2π mo ao                                                                         .                                             Comparing equs (3) and (4)  we get the same result for the Bohr's radius.                                              2π e^2/nKh  =  e^2/2nεo h =  h/n 2π mo ao

                                 or    ao  =   εo h^2 / π mo e^2

    By estimating the orbital velocity of the electron in the hydrogen atom and the corresponding change in its relativistic mass, one can show that the energy equivalent to the relativistic change of mass is equal to its binding energy with the nucleus.The velocity of the electron in the nth orbit in hydrogen atom is given by

vn = e^2/2nεo h =    (1.602 x 10^-19)^2 / 2n x (8.85 x 10^-12) x (6.626 x 10^-34) 

                    = 0.02188 x 10^8/n =[ 2.188 x 10^6 /n] m/s

 vn  =  h/n 2π mo ao =  7 x 6.626 x 10^-34 / n x 22 x 9.11 x 10^-31 x 5.29 x 10^-11

                                                   = [2.187 x 10^6 /n] m/s

vn^2/c^2  =  (2.188 x 10^6)^2/n^2 (2.998 x 10^8)^2 = 4.787344 x 10^12 / n^2 x 8.988004 x 10^16

                   = [0.5326370571/n^2] x 10^-4 = 0.00005326370571/n^2

The mass of the orbital electron in the nth orbit  mo /[1- vn^2/c^2]^1/2  ≃ mo[1 + vn^2/2c^2]

The relativistic increase of mass  Δm = mo vn^2/2c^2 and its equivalent energy  Δm c^2 = [mo vn^2//2]j  = [mo vn^2/2e ] eV

                                = 9.108 x 10^-31x(2.188 x 10^6)^2/ [2 x (1.602 x 10^ -19) n^2 

                                = 9.108 x 2.393672 x 10^-19 / n^2  j =  13.6 /n^2  eV

It is in agreement with the practical value of  ionization energy of hydrogen atom from its various energy levels. 

Bohr's Theory of Hydrogen atom -5

 Spectral feature in terms of matter wavelength of orbital electron 

     The electrons in atomic orbits behave like matter waves .It provides an acceptable explanation for the discreetness of the electronic orbits in atoms. When an electron jumps from outer orbit to any one the inner orbits, its radius and potential energy  are  decreased ,orbital velocity and binding energy are increased and these changes result with a decrease in the matter waves associated with the orbital electron.  During this jumping the excess energy is radiated out as electromagnetic radiation of wavelength  λwave,.  As they are all the after-effects of the above electronic transition, one may believe that they are linked indirectly. Based on this assumption, let us determine the wavelength of em radiation emitted in a transition from orbit with quantum number n2  to orbit with quantum number n1.

 

. Two electronic orbits with matter waves of orbital electron

   The energy of transition  (ΔE) from n2 to n1   = E2 - E1,  ,where E1,E2 denote the total energy of the orbital electron in the orbit 1 and 2 respectively.

                                       E2- E1 = ΔE = h ν = hc/ λwave  or λwave  =  hc/ΔE2→1

The wavelengths of matter wave associated with the electron in the orbit with quantum number n2 and n1  are h/mv2 and h/mv1 respectively. It gives  mv1 = h/λmatter1 and mv2   = h/λmatter2.  With this knowledge one can estimate the kinetic and potential energies of the orbital electron in terms of  its wavelength of matter wave. 

Kinetic energy of the electron in orbit n2 = (1/2) m v2^2 = (1/2m)  h^2/λ^2matter2.

Potential energy of the electron in orbit n2 = - e^2 /Kr2 = - m v2^2  = - (1/m) h^2/λ^2matter2.

Total energy of the electron in the orbit n2 = Sum of kinetic and potential energies E2 = - (1/2m)  h^2/λ^2matter2

Total energy of the electron in the orbit n1 = E1 =  - (1/2m) h2/λ^2matter1

 ΔE2→1 = (h^2/2m) [ 1/λ^2matter1 - 1/λ^2matter2 ]

The radius of the electronic orbit with quantum number n in hydrogen atom is shown as rn = n^2 ao , where ao is Bohr's radius, the radius of the innermost orbit of the electron in the hydrogen atom. From this one can derive mv^2 = e^2/ Kn^2ao and λ^2matter = h^2 Kn^2 ao /me^2 = (2πnao)^2 .  This can be verified by substituting the values for matter wavelength in  ΔE2→1

      ΔE2→1  = (h^2/2m) [1/(2πao)^2][1/n2^2 - 1/n1^2] = [h^2 /8π^2m ao^2][1/n2^2 - 1/n1^2]  = [e^4 m/8 h^2 εo^2] [1/n2^2 - 1/n1^2]

and  λ2→1 = 8 c εo^2 h^3 / (1/nf^2 - 1/ni^2)m e^4

Bohr's Theory of Hydrogen atom -4

 Energy of the orbital electron in various orbits of hydrogen atom

       When the electron is in the innermost orbit of the hydrogen atom, it is said to be in its ground state with lowest energy, when the electron is  in any allowed higher orbits it is said to be in its excited state with more energy. 

       Total energy of the electron  in a privileged orbit denoted by n is En.   

                En  = -  e^2/ (8 πεorn)

Substitute the value of rn  from   En  = - [e^2/(8 πεo)] x [mπe^2/n^2 εoh^2]                                                                                                          = - [e^4 x m]/8 εo^2h^2] x (1/n^2)                           

When the electron is in the first stable orbit  the system (n=1) is said to be in the ground state. The energy of the system  becomes E1  =  - [e^4 x m]/8 εo^2h^2]  

                = - [(1.602 x 10^-19)^4 x 9.108 x 10^-31]/[8 x (8.85 x 10^-12)^2 x(6.626x 10^-34 )^2]                    =  - (1.602)^4  x 9.108 x 10^-15 / 8 x (8.85)^2  x (6.626)^2                                                                        = - 2.18 x 10^-18  j =- 2.18  x 10^-18/1.602 x 10^-19   = -13.6 e.V

     When the electron is in the higher orbits with n ≥ 2, the system is said to be in the excited state. The system has less binding energy. The energy of the electron in the excited states En =  - [e^4 x m] / 8 εo^2h^2] x (1/n^2)  where n = 2,3,4..... or  En = - 13,6 / n^2   eV 

Electronic transition energy

     When the electron is in any one of the excited states, automatically without any external influence, it turns into its ground state, where the potential energy is minimum . During this transition, the difference in total energy is emitted out as em waves.

                        ΔE = E excited /initial  - E ground / final

                                          = [ e^4 x m] / 8 εo^2h^2] [(1/nf^2 - 1/ni^2)]                                                           

The frequency of the emitted radiation ν =  ΔE/h = [m e^4 /8 εo^2 h^3] [(1/nf^2 - 1/ni^2)]      

                The wavelength of the emitted radiation  λ = C/ ν=  8 c εo^2 h^3 / (1/nf^2 - 1/ni^2) m e^4        

 Radius of the permitted electronic orbits 

       It is governed by the relation  mvnrn = n[h/2π] where  h is  Planck's constant . Its value is 6.626 ×10−34 joule-seconds. Squaring both sides, m^2 vn^2 rn^2 = n^2 h^2/4 π^2  or vn^2  =   n^2h^2 /4π^2m^2  rn^2 .   Substituting this value in 

                     e^2/[4πεo] m    =  vn^2 rn =  n^2 h^2/4 π^2m^2  rn

                                         rn = [n^2 h^2/4 π^2 m^2] [4πεo m/ e^2 ] =   n^2 εoh^2/mπ e^2 = n^2 ao              It shows that the radius of the orbital electron is directly proportional to n^2 , that is why n is called as principle orbital quantum number. As we go away from the center the orbits are widely separated. For the innermost orbit  n=1, and r1 = ao , a constant called Bohr radius. The Bohr radius  is a fundamental constant in atomic physics, representing the most probable distance between the electron and the nucleus in the hydrogen atom in its ground state. The permitted electronic orbits in the hydrogen atoms have radii rn = n^2 ao. It shows that the radius of higher orbits   in hydrogen atom are 4 ao , 9 ao , 16 ao .....     n^2 ao .and  the spacing between the n th and (n+1) th orbits is (2n+1)ao .  

      The radius of the first orbit in the hydrogen atom is called Bohr radius and is given by   

                                  ao  =  h^2 εo /mπ e^2                                                                                                                                                                      

=[7 x (6.626 x 10^-34)2 x 8.85 x 10^-12]/[22 x (1.602 x 10^-19)2 x 9.108 x 10^-31]                              = 5.289 x 10^-11 m 

 Total energy of the orbital electron          

      The orbital electron has both kinetic energy due to its orbital motion and potential energy by virtue of its position in the electrostatic field. Kinetic energy of the single electron in the hydrogen atom = (1/2) m v^2. Since  e^2/[4πεo]r = mv^2 , K.E = (1/2)[e^2/[(4πεo)r] = e^2/8 πεor.  Potential energy of the electron = - e^2/4 πεor =  - 2 e^2/8 πεor .Total energy is

                                        TE= K.E + P.E = -   e^2/8πεor.                                                         

  It is negative and is responsible for its binding with the nucleus 

     Knowing the practical value of atomic binding energy of the electron in the innermost orbit of the hydrogen atom from the knowledge on its ionization energy, one can estimate Bohr radius. The validity of the theory can be verified by comparing the practical with the theoretical values. 

Binding energy =  13.6 eV  = 13.6 x 1.602 x 10^-19  = 2.1787 x 10^-18 j  ; [1 eV = 1.602 x 10^-19 joule]                e^2/8 πεor = 13.6 eV = 2.1787 x 10^-18 j

or         r = [1.602 x 10^-19]^2 / 8 x (22/7) x 8.85 x 10^-12 x 2.1787 x 10^-18                                                                                                                                                                                                              = [7 x (1.602)^2  x 10^-38] / 8 x 22 x 8.85 x 2.1787 x 10^-30                                                                   = [17.964828/3393.54312] x 10^-8                                                                                                               = 5.293 x 10^-11 m                                                                                                                                              It is very same as that of Bohr radius derived earlier.

 Bohr's Theory of Hydrogen atom

The first Bohr's postulate  states that  the  angular momentum  of the orbital electron is some integral multiples of h/2π.    mvn rn = nh/ 2π

                                            or    2πrn  = n[h/mvn] = nλ

i.e., the circumference of all the  allowed electronic orbits in the hydrogen atom must accommodate some integral number of a kind of wave characterizing the orbital electron - matter wave associated with the electron. It implies that                                                                                  

                                               h/2πm =  vnrn /n = constant                                                       

The wave characteristics of the electron make the electronic orbits to be discrete.

     For stability of the orbital electron in the Hydrogen atom, the electro-static force of attraction between the nucleus and the electron in the circular orbit  must be equal to the centrifugal force at every instant of its motion.  

                             electrostatic force of attraction         =  centrifugal force

                                               e^2/[4πεo] rn^2                          =  mvn^2/rn                                      

                                                e^2/[4πεo] m     = constant    =  vn^2 rn

These two conditions explain the dependence of one variable with the other used in Bohr's theory of hydrogen atom.  

(1) Orbital velocity (vn) Vs Orbital radius (rn) 

     The equation predicts  vn^2rn  = constant i.e, vn is inversely proportional to rn^(1/2)  i.e., as rn increases the orbital velocity decreases. By substituting rn = n^2 ao  we get  vn^2 n^2 ao  = constant , where ao is Bohr radius, the radius of the innermost orbit in hydrogen atom.  

(2) Orbital velocity (vs) Vs Quantum number(n)

    From the above equations one can derive  vn^2 rn =  e^2/ m[4πεo] = constant and  vnrn/n  =  [h/2πm] = constant. Dividing one by the other we have   vn  = [e^2/2n εo h] ,  or nvn = constant ,which indicates that vn is inversely proportional to n.i.e, as we go away from the center ,the orbital electron moves slower.The very same conclusion can be obtained from the equation  vnrn /n =  vn^2 rn / vnn  = constant. since  vn2rn  = constant , vn n must be a constant. Again from the  relation  mvnrn   = nh/2π,   vn rn /n  = constant.  vn rn /n  =  vn^2rn /nvn  or nvn  = constant  as vn2 rn  = constant . 

    One can establish the same relationship between vn and n from the Bohr's condition on the circumference of the electronic orbits. All orbits irrespective of its quantum number accommodate some integral number of matter waves associated with the orbital electron 

                                                              2πrn = nλ = nh/m vn  

                                                            2πm/h = n/vnrn = constant

Since vn^2rn  = constant  nvn = constant

(3) Orbital readius (rn) Vs Quantum number (n)

     The relation  mvnrn   = nh/2π   gives  vnrn/n  = constant.  vn rn /n  =  vn^2 rn /n vn  or n vn  = constant  as vn^2rn  = constant . On Squaring the relation  [vn rn /n]2  = vn^2 rn  /[n^2 /rn] or  n^2 /rn  = constant as vn^2 rn  = constant. It shows that rn  is directly proportional to n^2. By substituting the value of rn  , we get   vn^2 n^2 ao = constant

 (4) Rate of variation of orbital velocity with respect to radius (dvn/drn)

          Differentiating  v^2 r = constant with respect to r, v^2 + 2vr (dv/dr) = 0, which gives  dvn/drn = -vn/2rn. By substituting the values of rn  and the corresponding vn  one can find out the rate with which the orbital velocity falls with radius of the orbit.

rn =  n^2 h^2 εo /mπ e^2   = n^2  ao

mvnrn  =  mvn(n^2)ao =  nh/2π  or  vn  =   h/2π mnao  which states nvn = constant 

Substituting these values in dvn/drn  we arrive at   dvn/drn = - vn/2rn   = -  h/4πmao^2 n^3. i.e.,  dvn/drn  is negative and varies inversely proportional to n3.