The Line Spectrum of Hydrogen

Hydrogen spectrum:

Let us consider the energy changes when an electron in higher energy level say n₂ jumps to lower energy level n₁.Whenever such a transition takes place radiations are emitted corresponding to the difference in the two  energy levels.

 ΔE = E₂ - E1 = 2π²mZ²e⁴k²/h²[ 1/n₁² - 1/n₂²]

                      = (IE)_Z [ 1/n₁² - 1/n₂²]

                     = (IE)_HZ²[ 1/n₁² - 1/n₂²]

Where, symbols have their usual meanings,  (IE)_H being the Ionization energy of hydrogen and (IE)_z  is the Ionization energy of an  the atom with atomic number.

Now, we know that ΔE  =  hc / λ

 ? wave number    = ΔE / hc

                                = 2π²mZ²e⁴k⁴/ch³[ 1/n₁² - 1/n₂²]

                  =  R_HZ²[ 1/n₁² - 1/n₂²].  Where, RH is the Rydbergsinglequotes Constant.

                               = (IE)_HZ²/ hc[ 1/n₁² - 1/n₂²]

The value of Rydbergsinglequotes constant R_H = 2π²mZ²e⁴k⁴/ch³ =  1.09678 X 10⁷ m^-1

Frequency ν = c / λ = velocity / wavelength = velocity X wave number = R_HZ²c[1/n₁² - 1/n₂²].

The following figure depicts the spectrum of hydrogen atom for various transition of the elctron from one energy level to another.

Thus, lyman series is the series for which n₁ is always 1 , balmer is the series for which n₁ is always 2  and so on.

The lines of the lyman series belong to the ultraviolet region , the lines in the balmer series belong to the visible region while lines of the other regions belong to the infrared regions.

The lines in balmer series are visible.

As, we go on increasing the value of  n₂ in each series , the series converges beyond a point and the spectrum becomes continuous.This corresponds to the case when the value of n2 approaches ? .This also gives the spectral line of maximum wavenumber or minimum wavelength for  that series.

That is  ,_max  = 1/ λ_min =  R_HZ²/n₁²

Also, the way we got maximum wave number when n₂ ? ? , minimum wave number is obtained when the difference [1/n₁² - 1/n₂²] is minimum, i.e., n₂ = n₁ + 1.

So , _min  = 1/ λ_max =  R_HZ²/[1/n₁² - 1/(n₁+1)²]

This corresponds to the first line of series (starting with the minimum wave number) , and is also called α line of the series.

Similarly , second, third , fourth lines etc. are called β,γ,δ lines of the series respectively.

As , the value of n₁ increases , that is we move away from the nucleus the spectral become more densely packed i.e, they come closer to each other and finally merge to form a continuous spectrum, that is there is no line structure but a continuous patch of emission or absorbtion spectrum.The reason for such behaviour is that once the electron is free from the attractive force of the nucleus its energy is no longer is quantized but it can assume absolutely any value. Since, it is under no force field so the potential energy is zero and the only energy is kinetic energy which is 1/2mv² , now the velocity in free space could be anything so energy can also be anything.

Merits and demerits of the Bohrsinglequotes Atomic Model:

Like the Rutherfordsinglequotes model of atom we discussed before Bohrsinglequotes model explained few phenomena successfully and had drawbacks in certain areas.

Bohrsinglequotes model successfully explained:

The stability of atoms.

The line spectrum of hydrogen was expalined satisfactorily.

It helped in understanding of related phenomena like emission of characteristics X-rays by elements.Experimental values satisfied the theoretical results for X-rays, Rydbergsinglequotes contant, excitation potential etc.

But , it suffered from some drawbacks:

• It was not able to explain the line spectra of multi-electron atoms. Though, the Bohrsinglequotes model successfully explained the line spectra of hydrogen like particles which were single electron entities but it failed to explain the spectra of multi electron atoms.With spectroscopes with better resolving powers were used , it was observed that even in case of hydrogen each lines were split up into a number of closely spaced lines (called fine structure). Bohr model could not explain this phenomenon.

• In the presence of strong electric field or magnetic field , each line was split was into a number of closey spaced line (called Starksinglequotes effect and Zeemansinglequotes effect respectively).Bohrsinglequotes model could not explain these phenomena.

• Inability in expalining the three dimensional model of atom. According to Bohrsinglequotes theory , electrons moved in two dimensional planetary orbits.This gave the flat model of the atom, but the atom is three dimensional and not flat.

• Inability to explain shapes of moleculesm ,like covalent molecules possessed different shapes unexpalined by Bohrsinglequotes model.

• Inability to explain the de broglie dual nature of matter and heisenbergsinglequotes uncertainty principle.

 Sommerfeld Extension: The fine structure observed in the atomic spectrum of hydrogen was against the Bohrsinglequotes theory of fixed energy levels for the electrons.According to Bohr , if an electron does a transition from one energy level to another , the difference in energy is emiited as a single line of that series.But , the fine structure suggest that the frequency may vary slightly for each series i.e., for each pair of energy levels; which indicates the  presence of energy sublevel (we will call it azimuthal quantum number, it will taken up later) for each orbit of Bohrsinglequotes energy level.

Somerfeld suggested that the electrons travel in elliptical orbit rather than circular orbits around a nucleus and corresponding to each principal quantum number (n) , several elliptical orbits were possible and the nucleus was located at the focus of the focus of the ellipse describing the electronsinglequotes orbit.

Hence, the motion of the electron was defined by two quantum numbers in an elliptical orbit :

• radical quantum number n_r and
  
• azimuthal quantum number k such that ,

                                                      n = n_r + k

and, n/ k = (length of semi-major axis of the ellipse)/ (length of semi minor axis of the ellipse)

Also, the ellipicity of the axis was calcualted by the relative values of n and k. For a particular of n , k = 1,2,3 ...................n .

Thus, when n = k , the orbit was cicular and otherwise the orbit was elliptical.

Thus, when n = 3, following orbits are possible:

 According to theory of relativity , the mass of a particles moving with speeds comparable to speed of light, is different to its rest mass.Also, the velocity of electron keeps changing in an elliptical orbit , its magnitude increases when it  approaches the nucleus and it decreases when it gets away.The expression of variation of mass with velocity is given by,

                               m = m₀/?(1 - v²/c²) , where m0 = mas of the electron at rest.

Sommerfeld, showed that the variation is mass with velocity caused the energy to vary slightly in orbits of varying ellipticity.And, the electron no longer travelled in a fixed closed path but and the semi-major axis shifted gradually.

This, explained the fine structure obtained as the electrons underwent transition from one set of values of n,k to another.But , only those transitions were allowed for which the change in the value of azimuthal quantum number was not more than one unit. Below is the figure for n = 5. ( k is replaced by l ).

                                                                                                                                            The elliptical orbits for n = 5 

   From Bohrsinglequotes theory, energy E_n is given by ,

                                      E_n = - 2π²me⁴Z²k²/ n²h²   ............(i)

From Sommerfeldsinglequotes theory ,

                                                   E_n = - 2π²me⁴Z²k²/ (n_r+k)²h² ............(ii)

So, according to sommerfeld, the energy of the electron depended on the eccenctricity of the ellipse.So, as per the theory the velocity of the electron varied at different distances to keep the electrostatic forces balanced.

[SolvedExample]

Example2:L Calculate the energy of electron in the first orbit of hydrogen.

Solution:E_n = 2π²me⁴Z²k²/ n²h²

Now, for hydrogen, Z = 1 , n= 1 too as we talking about the first orbit.

In CGS units, k =1. m = 9 X 10^-28 g , e = 4.8 X 10^-10 esu.

So, E₁ = - 2 X (3.14)² X 9 X 10^-28 X (4.8 X 10^-10)²/(6.63 X 10^-27)²  / (6.63 X 10^-27)²

              = -2.18 X 10^-11 erg.

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[SolvedExample]

Example3: Suppose the Ionization energy of Li²⁺ is 4.4 X 10^-17 J. Calulate the energy of first energy state of He⁺.

For lithium , Z =3.

The ionization energy of first orbit of Li²⁺ is 4.4 X 10^-17 J. (It is IE of Li²⁺ and not Li atom).

So, the IE of hydrogen atom = 4.4 X 10^-17 / 9  J

IE of He+  = 4 X  4.4 X 10^-17 / 9 J = 19.6 x 10^-18 J.

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[SolvedExample]

Example4: Calculate the wavelength when an electron jumps from second orbit to first orbit in hydrogen atom. Rydberg’s constant R = 109737cm-1

Solution: For hydrogen, Z =1.

As per the Rydberg’s formula 1/ λ = R( 1/n₁² – 1/n₂²).

                                                       =  109737 (1 – ¼) cm

                                                      = 109737 X ¾  cm

                                                      = 82302.75 cm/wave

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[SolvedExample]

Example 5: What electronic transition in hydrogen atom would emit the radiation of same wavelength as the second Balmer series transition for He⁺ ion? Also find second ionization potential of He and first Bohr orbit for He⁺.

Given, R (for both He⁺ and H)  is 109737cm^-1.

(e = 1.6 X 10^-19 coulomb, m = 9.1X 10^-31 Kg , h = 6.626 X 10^-34 J.s , c = 3.8 X 10⁸ m/sec , ε = 8.854 X 10^-12 coulomb

Solution: For second line of Balmer series for He+  we have, n₁ =2, n₂ = 4 (Z =2).

Therefore, by Rydberg’s formula we have,

1/λ = RZ² (1/4 -1/16)

1/λ = 3R/4

For the same transition in Hydrogen atom we have,

3R/4 = R (1/n₁² -1/n₂²)

It is only possible for n₁ =1 and n₂ = 2.

Hence, it is  the first line of Lyman series of hydrogen.

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