Mathematics THE LINE SPECTRA OF THE HYDROGEN ATOM AND DE BROGLIE’S EXPLANATION OF BOHR’S SECOND POSTULATE OF QUANTISATION

### Topic Covered

color{blue}{star} THE LINE SPECTRA OF THE HYDROGEN ATOM
color{blue}{star} DE BROGLIE’S EXPLANATION OF BOHR’S SECOND POSTULATE OF QUANTISATION

### THE LINE SPECTRA OF THE HYDROGEN ATOM

color{blue} ✍️According to the third postulate of Bohr’s model, when an atom makes a transition from the higher energy state with quantum number ni to the lower energy state with quantum number n_f " "(n_f < n_i), the difference of energy is carried away by a photon of frequency ν_(if) such that hv_(if) = En_i -En_f.... (12.20)

color{blue} ✍️Using Eq. (12.16), for En_f and En_i, we get

color{navy}(hv_(if)= (me^4)/(8ε_(0)^(2)h^3) (1/n_(f)^(2)-1/n_(1)^(2))

...................(12.21)

or

color{navy}(v_(if)= (me^4)/(8ε_(0)^(2)h^3) (1/n_(f)^(2)-1/n_(1)^(2))

.................(12.22)

color{blue} ✍️Equation (12.21) is the Rydberg formula, for the spectrum of the hydrogen atom. In this relation, if we take n_f = 2 and n_i = 3, 4, 5..., it reduces to a form similar to Eq. (12.10) for the Balmer series. The Rydberg constant R is readily identified to be

color{navy}(R= (me^4)/(8ε_(0)^(2)h^3C))

................. (12.23)

color{blue} ✍️If we insert the values of various constants in Eq. (12.23), we get R = 1.03 × 10^7 m^(–1)

color{blue} ✍️This is a value very close to the value color{purple}((1.097 × 10^7 m^(–1)) obtained from the empirical Balmer formula.
This agreement between the theoretical and experimental values of the Rydberg constant provided a direct and striking confirmation of the Bohr’s model.

color{blue} ✍️Since both n_f and n_i are integers, this immediately shows that in transitions between different atomic levels, light is radiated in various discrete frequencies. For hydrogen spectrum, the Balmer formula corresponds to n_f = 2 and n_i = 3, 4, 5, etc.

color{blue} ✍️The results of the Bohr’s model suggested the presence of other series spectra for hydrogen atom–those corresponding to transitions resulting from n_f = 1 and n_i = 2, 3, etc.; n_f = 3 and n_i = 4, 5, etc., and so on. Such series were identified in the course of spectroscopic investigations and are known as the L yman, Balmer, Paschen, Brackett, and Pfund series. The electronic transitions corresponding to these series are shown in Fig. 12.9.

color{blue} ✍️The various lines in the atomic spectra are produced when electrons jump from higher energy state to a lower energy state and photons are emitted. These spectral lines are called emission lines.

color{blue} ✍️But when an atom absorbs a photon that has precisely the same energy needed by the electron in a lower energy state to make transitions to a higher energy state, the process is called absorption.

color{blue} ✍️Thus if photons with a continuous range of frequencies pass through a rarefied gas and then are analysed with a spectrometer, a series of dark spectral absorption lines appear in the continuous spectrum. The dark lines indicate the frequencies that have been absorbed by the atoms of the gas.

color{blue} ✍️The explanation of the hydrogen atom spectrum provided by Bohr’s model was a brilliant achievement, which greatly stimulated progress towards the modern quantum theory. In 1922, Bohr was awarded Nobel b Prize in Physics.
Q 3119567419

Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.
Class 12 Chapter 12 Example 6
Solution:

The Rydberg formula is

hc//lamda_(if) = (me^4)/(8ε_(0)^(2)h^2) (1/n_(f)^(2) - 1/n_(1)^(2))

The wavelengths of the first four lines in the Lyman series correspond to transitions from n_i = 2,3,4,5 to n_f = 1. We know that

(me^4)/(8ε_(0)^(2)h^2) = 13.6 eV = 21.76 ×10^(–19) J

Therefore,

lamda_(if) = (hc)/(21.76xx10^(-19) (1/1-1/n_(1)^(2))m
= (6.625xx10^(-34)xx3xx10^8xxn_(1)^(2))/(21.76xx10^(-19) xx(n_(1)^(2)-1))m = (0.9134n_(1)^(2))/(n_(1)^(2)-1)xx10^(-7)m

= 913.4 n_(i)^(2)//(n_(i)^(2) –1) Å

Substituting n_i = 2,3,4,5, we get λ_(21) = 1218 Å, λ_(31) = 1028 Å, λ_(41) = 974.3 Å, and λ_(51) = 951.4 Å.

### DE BROGLIE’S EXPLANATION OF BOHR’S SECOND POSTULATE OF QUANTISATION

color{blue} ✍️All of the postulates, Bohr made in his model of the atom, perhaps the most puzzling is his second postulate. It states that the angular momentum of the electron orbiting around the nucleus is quantised (that is, L_n = nh//2π; n = 1, 2, 3 …)

color{blue} ✍️We studied, about the de Broglie’s hypothesis that material particles, such as electrons, also have a wave nature. C. J. Davisson and L. H. Germer later experimentally verified the wave nature of electrons in .

color{blue} ✍️Louis de Broglie argued that the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave. In analogy to waves travelling on a string, particle waves too can lead to standing waves under resonant conditions.

color{blue} ✍️We know that when a string is plucked, a vast number of wavelengths are excited. However only those wavelengths survive which have nodes at the ends and form the standing wave in the string.

color{blue} ✍️It means that in a string, standing waves are formed when the total distance travelled by a wave down the string and back is one wavelength, two wavelengths, or any integral number of wavelengths.

color{blue} ✍️Waves with other wavelengths interfere with themselves upon reflection and their amplitudes quickly drop to zero. For an electron moving in nth circular orbit of radius r_n, the total distance is the circumference of the orbit, 2πrn. Thus

color{navy}(2π r_n = nλ, n = 1, 2, 3...)

......... (12.24)

color{blue} ✍️Figure 12.10 illustrates a standing particle wave on a circular orbit for n = 4, i.e., 2πrn = 4λ, where λ is the de Broglie wavelength of the electron moving in nth orbit. From Chapter 11, we have λ = h/p, where p is the magnitude of the electron’s momentum. If the speed of the electron is much less than the speed of light, the momentum is mv_n. Thus, λ = h//mv_n. From Eq. (12.24), we have

color{purple}(2π r_n = n h//mv_n" or "m v_n r_n = nh//2π)

color{blue} ✍️This is the quantum condition proposed by Bohr for the angular momentum of the electron [Eq. (12.13)]. In Section 12.5, we saw that this equation is the basis of explaining the discrete orbits and energy levels in hydrogen atom.

color{blue} ✍️Thus de Broglie hypothesis provided an explanation for Bohr’s second postulate for the quantisation of angular momentum of the orbiting electron. The quantised electron orbits and energy states are due to the wave nature of the electron and only resonant standing waves can persist.

color{blue} ✍️Bohr’s model, involving classical trajectory picture (planet-like electron orbiting the nucleus), correctly predicts the gross features of the hydrogenic atoms*, in particular, the frequencies of the radiation emitted or selectively absorbed. This model however has many limitations. Some are:

color{blue} {(i)} The Bohr model is applicable to hydrogenic atoms. It cannot be extended even to mere two electron atoms such as helium. The analysis of atoms with more than one electron was attempted on the lines of Bohr’s model for hydrogenic atoms but did not meet with any success. Difficulty lies in the fact that each electron interacts not only with the positively charged nucleus but also with all other electrons.

color{blue} ✍️The formulation of Bohr model involves electrical force between positively charged nucleus and electron. It does not include the electrical forces between electrons which necessarily appear in multi-electron atoms.

color{blue} {(ii)} While the Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms, the model is unable to explain the relative intensities of the frequencies in the spectrum.

color{blue} ✍️In emission spectrum of hydrogen, some of the visible frequencies have weak intensity, others strong. Why? Experimental observations depict that some transitions are more favoured than others. Bohr’s model is unable to account for the intensity variations.

color{blue} ✍️Bohr’s model presents an elegant picture of an atom and cannot be generalised to complex atoms. For complex atoms we have to use a new and radical theory based on Quantum Mechanics, which provides a more complete picture of the atomic structure.