Mathematics ATOMIC SPECTRA AND BOHR MODEL OF THE HYDROGEN ATOM

### Topic Covered

color{blue}{star} ATOMIC SPECTRA
color{blue}{star} BOHR MODEL OF THE HYDROGEN ATOM

### ATOMIC SPECTRA

color{blue} ✍️As we know, each element has a characteristic spectrum of radiation, which it emits. When an atomic gass or vapor is excited at low pressure, usually by passing an electric current through it, the emitted radiation has a spectrum which contains certain specific wavelengths only.

color{blue} ✍️A spectrum of this kind is termed as emission line spectrum and it consists of bright lines on a dark background. The spectrum emitted by atomic hydrogen is shown in Fig. 12.5.

color{blue} ✍️Study of emission line spectra of a material can therefore serve as a type of “fingerprint” for identification of the gas. When white light passes through a gas and we analyse the transmitted light using a spectrometer we find some dark lines in the spectrum.

color{blue} ✍️These dark lines correspond precisely to those wavelengths which were found in the emission line spectrum of the gas. This is called the absorption spectrum of the material of the gas.

color{brown}bbul("Spectral series")
color{blue} ✍️We might expect that the frequencies of the light emitted by a particular element would exhibit some regular pattern. Hydrogen is the simplest atom and therefore, has the simplest spectrum.

color{blue} ✍️In the observed spectrum, however, at first sight, there does not seem to be any resemblance of order or regularity in spectral lines. But the spacing between lines within certain sets of the hydrogen spectrum decreases in a regular way (Fig. 12.5).

color{blue} ✍️Each of these sets is called a spectral series. In 1885, the first such series was observed by a Swedish school teacher Johann Jakob Balmer in the visible region of the hydrogen spectrum. This series is called Balmer series (Fig. 12.6).

color{blue} ✍️The line with the longest wavelength, 656.3 nm in the red is called Hα; the next line with wavelength 486.1 nm in the bluegreen is called Hβ, the third line 434.1 nm in the violet is called Hγ; and so on.

color{blue} ✍️As the wavelength decreases, the lines appear closer together and are weaker in intensity. Balmer found a simple empirical formula for the observed wavelengths

color{navy}(1/lamda = R (1/(2^2) - 1/(n^2)))

......(12.5)

where λ is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc.
The value of R is color{purple}(1.097 × 10^7 m^(–1)). This equation is also called Balmer formula.

"Paschen series:"

color{blue} ✍️Taking n = 3 in Eq. (12.5), one obtains the wavelength of the Hα line:

color{purple}(1/lamda = 1.097xx 10^7( 1/(3^2) - 1/(n^2))

color{purple}(=1.522 xx10^6m^(-1))

color{purple}(i.e., lamda= 656.3nm)

color{blue} ✍️For n = 4, one obtains the wavelength of H_β line, etc. For n = ∞, one obtains the limit of the series, at λ = 364.6 nm. This is the shortest wavelength in the Balmer series. Beyond this limit, no further distinct lines appear, instead only a faint continuous spectrum is seen. Other series of spectra for hydrogen were subsequently discovered. These are known, after their discoverers, as Lyman, Paschen, Brackett, and Pfund series. These are represented by the formulae:

color{brown}bbul("Lyman series:")

color{navy}(1/lamda = R (1/(1^2) - 1/(n^2)) \ \ \ \ n = 2,3,4)

......... (12.6)

color{brown}bbul("Paschen series:")

color{blue}(1/lamda = R (1/(3^2) - 1/(n^2)) \ \ \ \ n = 4,5,6)

......... (12.7)

color{brown}bbul("Brackett series:")

color{blue}(1/lamda = R(1/(4^2)-1/(n^2)) \ \ \ n = 5,6,7)

........... (12.8)

color{brown}bbul("Pfund series:")

color{navy}(1/lamda = R(1/(5^2)-1/(n^2)))

............ (12.9)

color{blue} ✍️The Lyman series is in the ultraviolet, and the Paschen and Brackett series are in the infrared region. The Balmer formula Eq. (12.5) may be written in terms of frequency of the light, recalling that

color{purple}(c = v lamda)

or color{purple}(1/lamda= v/c)

color{blue} ✍️Thus, Eq. (12.5) becomes

color{blue}(v = Rc (1/2^2-1/n^2))

.... (12.10)

color{blue} ✍️There are only a few elements (hydrogen, singly ionised helium, and doubly ionised lithium) whose spectra can be represented by simple formula like Eqs. (12.5) – (12.9).

color{blue} ✍️Equations (12.5) – (12.9) are useful as they give the wavelengths that hydrogen atoms radiate or absorb. However, these results are empirical and do not give any reasoning why only certain frequencies are observed in the hydrogen spectrum.

### BOHR MODEL OF THE HYDROGEN ATOM

color{blue} ✍️The model of the atom proposed by Rutherford assumes that the atom, consisting of a central nucleus and revolving electron is stable much like sun-planet system which the model imitates.

color{blue} ✍️However, there are some fundamental differences between the two situations. While the planetary system is held by gravitational force, the nucleus-electron system being charged objects, interact by Coulomb’s Law of force.

color{blue} ✍️We know that an object which moves in a circle is being constantly accelerated – the acceleration being centripetal in nature. According to classical electromagnetic theory, an accelerating charged particle emits radiation in the form of electromagnetic waves. The energy of an accelerating electron should therefore, continuously decrease. The electron would spiral inward and eventually fall into the nucleus (Fig. 12.7).

color{blue} ✍️Thus, such an atom can not be stable. Further, according to the classical electromagnetic theory, the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of revolution.

color{blue} ✍️As the electrons spiral inwards, their angular velocities and hence their frequencies would change continuously, and so will the frequency of the light emitted. Thus, they would emit a continuous spectrum, in contradiction to the line spectrum actually observed. Clearly Rutherford model tells only a part of the story implying that the classical ideas are not sufficient to explain the atomic structure.
Q 3159267114

According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.
Class 12 Chapter 12 Example 4
Solution:

From Example 12.3 we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10–11 m is 2.2 × 10–6 m//s. Thus, the frequency of the electron moving around the proton is

v=v/(2pir) = (2.2xx10^6 ms^(-1))/(2pi(5.3xx10^(-11)m)

≈ 6.6 × 10^(15) Hz.

According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 10^(15) Hz.
color{blue} ✍️It was Niels Bohr who made certain modifications in this model by adding the ideas of the newly developing quantum hypothesis. Niels Bohr studied in Rutherford’s laboratory for several months in 1912 and he was convinced about the validity of Rutherford nuclear model.

color{blue} ✍️Faced with the dilemma as discussed above, Bohr, in 1913, concluded that in spite of the success of electromagnetic theory in explaining large-scale phenomena, it could not be applied to the processes at the atomic scale. It became clear that a fairly radical departure from the established principles of classical mechanics and electromagnetism would be needed to understand the structure of atoms and the relation of atomic structure to atomic spectra.

color{blue} ✍️Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates. These are :

color{blue} {(i)} Bohr’s first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy, contrary to the predictions of electromagnetic theory. According to this postulate, each atom has certain definite stable states in which it can exist, and each possible state has definite total energy. These are called the stationary states of the atom.

color{blue} {(ii)} Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/(2π) where h is the Planck’s constant (= 6.6 × 10^(–34) J s).

color{blue} ✍️Thus the angular momentum (L) of the orbiting electron is quantised. That is

color{blue} {L = nh//2π}

........ (12.11)

color{blue} {(iii)} Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by

color{blue} {hν = E_i – E_f}

......(12.12)

color{blue} ✍️where E_i and E_f are the energies of the initial and final states and E_i > E_f . For a hydrogen atom, Eq. (12.4) gives the expression to determine the energies of different energy states. But then this equation requires the radius r of the electron orbit. To calculate r, Bohr’s second postulate about the angular momentum of the electron–the quantisation condition – is used. The angular momentum L is given by color{purple} {L = mvr}

color{blue} ✍️Bohr’s second postulate of quantisation [Eq. (12.11)] says that the allowed values of angular momentum are integral multiples of h//2π.

color{blue}(L_n = mv_nr_n = (nh)/(2pi))

........(12.13)

color{blue} ✍️where n is an integer, r_n is the radius of nth possible orbit and n^th is the speed of moving electron in the nth orbit. The allowed orbits are numbered 1, 2, 3 ..., according to the values of n, which is called the principal quantum number of the orbit.
From Eq. (12.3), the relation between v_n and r_n is

color{purple}(v_n = e/sqrt(4piε_0mr_n))

color{blue} ✍️Combining it with Eq. (12.13), we get the following expressions for v_n and r_n,

color{blue}(v_n = 1/n (e^2)/(4piε_0) 1/(h/22pi))

..............(12.14)

and

color{blue}(r_n = ((n^2)/m) (h/(2pi))^2 (4piε_0)/(e^2))

.............. (12.15)

color{blue} ✍️Eq. (12.14) depicts that the orbital speed in the nth orbit falls by a factor of n. Using Eq. (12.15), the size of the innermost orbit (n = 1) can be obtained as

color{purple}(r_1 = (h^2ε_0)/(pime^2))

color{blue} ✍️This is called the Bohr radius, represented by the symbol a_0. Thus,

color{blue}(a_0 = (h^2ε_0)/(pime^2))

.....(12.16)

Substitution of values of h, m, ε0 and e gives color{purple}(a0 = 5.29 × 10^(–11)) m. From Eq. (12.15), it can also be seen that the radii of the orbits increase as n^2.

color{blue} ✍️The total energy of the electron in the stationary states of the hydrogen atom can be obtained by substituting the value of orbital radius in Eq. (12.4) as

color{purple}(E_n = - (e^2)/(8piε_0) (m/(n^2)) ((2pi)/h)^2 (e^2)/(4piε_0))

or

color{blue}(E_n = - (me^4)/(8n^2ε_(0)^(2)h^2))

............ (12.17)

color{blue} ✍️Substituting values, Eq. (12.17) yields

color{blue}(E_n = (2.18xx10^(-18))/((n^2) J)

........ (12.18)

color{blue} ✍️Atomic energies are often expressed in electron volts (eV) rather than joules. Since color{purple} {1 eV = 1.6 × 10^(–19) J}, Eq. (12.18) can be rewritten as

color{blue}(E_n = - (13.6)/(n^2) eV)

..... (12.19)

color{blue} ✍️The negative sign of the total energy of an electron moving in an orbit means that the electron is bound with the nucleus. Energy will thus be required to remove the electron from the hydrogen atom to a distance infinitely far away from its nucleus (or proton in hydrogen atom).

color{blue} ✍️The derivation of Eqs. (12.17) – (12.19) involves the assumption that the electronic orbits are circular, though orbits under inverse square force are, in general elliptical. (Planets move in elliptical orbits under the inverse square gravitational force of the sun.) However, it was shown by the German physicist Arnold Sommerfeld (1868 – 1951) that, when the restriction of circular orbit is relaxed, these equations continue to hold even for elliptic orbits.

Q 3139567412

A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite.
Class 12 Chapter 12 Example 5
Solution:

From Eq. (12.13), we have

m v_n rn = n+h//2π

Here m = 10 kg and r_n = 8 × 10^6 m. We have the time period T of the
circling satellite as 2 h. That is T = 7200 s.
Thus the velocity v_n = 2π rn//T.
The quantum number of the orbit of satellite
n = (2π r_n)^2 × m//(T × h).
Substituting the values,
n = (2π × 8 × 10^6 m)^2 × 10//(7200 s × 6.64 × 10^(–34) J s)
= 5.3 × 10^(45)
Note that the quantum number for the satellite motion is extremely large! In fact for such large quantum numbers the results of quantisation conditions tend to those of classical physics.

### Energy levels

color{blue} ✍️The energy of an atom is the least (largest negative value) when its electron is revolving in an orbit closest to the nucleus i.e., the one for which n = 1. For n = 2, 3, ... the absolute value of the energy E is smaller, hence the energy is progressively larger in the outer orbits.

color{blue} ✍️The lowest state of the atom, called the ground state, is that of the lowest energy, with the electron revolving in the orbit of smallest radius, the Bohr radius, a_0. The energy of this state (n = 1), E_1 is –13.6 eV.

color{blue} ✍️Therefore, the minimum energy required to free the electron from the ground state of the hydrogen atom is 13.6 eV. It is called the "ionisation energy" of the hydrogen atom.

color{blue} ✍️This prediction of the Bohr’s model is in excellent agreement with the experimental value of ionisation energy. At room temperature, most of the hydrogen atoms are in ground state. When a hydrogen atom receives energy by processes such as electron collisions, the atom may acquire sufficient energy to raise the electron to higher energy states.

color{blue} ✍️The atom is then said to be in an excited state. From Eq. (12.19), for n = 2; the energy E_2 is –3.40 eV. It means that the energy required to excite an electron in hydrogen atom to its first excited state, is an energy equal to E_2 – E_1 = –3.40 eV – (–13.6) eV = 10.2 eV.

color{blue} ✍️Similarly, E_3 = –1.51 eV and E_3 – E_1 = 12.09 eV, or to excite the hydrogen atom from its ground state (n = 1) to second excited state (n = 3), 12.09 eV energy is required, and so on.

color{blue} ✍️From these excited states the electron can then fall back to a state of lower energy, emitting a photon in the process. Thus, as the excitation of hydrogen atom increases (that is as n increases) the value of minimum energy required to free the electron from the excited atom decreases.

color{blue} ✍️The energy level diagram for the stationary states of a hydrogen atom, computed from Eq. (12.19), is given in Fig. 12.8. The principal quantum number n labels the stationary states in the ascending order of energy.

color{blue} ✍️In this diagram, the highest energy state corresponds to n =∞ in Eq, (12.19) and as an energy of 0 eV. This is the energy of the atom when the electron is completely removed (r = ∞) from the nucleus and is at rest. Observe how the energies of the excited states come closer and closer together as n increases.