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Physics Topics are also essential for space exploration, allowing scientists to study phenomena such as gravitational waves and cosmic rays.

Line spectrum: If electric discharge occurs inside any elementary gas or vapour kept in a discharge tube at a few mm of mercury pressure, the tube acts as a bright source of light. It is generally called discharge lamp. Discharge lamps of neon, sodium, mercury and halogen gases are used in our daily life.

With help of prism or other instruments different fundamental colours of different wavelengths are obtained in dispersion of light emitted from a discharge lamp. This is known as atomic spectrum. In this spectrum, there are ultraviolet rays and infrared rays along with visible light. With the help of special experimental arrangement, wavelength of each fundamental ray can be determined. Atomic spectra for different elements are different.

The characteristics of the atomic spectra are that these are usually a combination of some thin, bright and discrete lines; i.e., in between two bright lines there is a dark space. This spectrum is known as line spectrum [Fig.(a)]. On the other hand, the spectrum obtained from a heated solid (e.g., an incandescent tungsten lamp) is a continuous spectrum; the different coloured lights present in it form continuous illumination on the screen and there is no dark space in between them [Fig.(b)]. Again, molecular spectrum is usually a band spectrum. Instead of discrete lines, closely spaced groups of lines are so formed that each group appears to be a band. Between two consecutive bands, there is a dark space.

Balmer series of hydrogen spectrum: In the visible region of the atomic spectrum of hydrogen, there are four bright lines. The experimental values of their wavelengths are 6563Å, 4861Å, 4341Å and 4102Å. These four lines constitute the Balmer series of hydrogen spectrum. A swiss mathematics teacher, Balmer tried to express these wave lengths by a simple relation in 1884, many years before the proposal of Bohr model, which is \(\frac{1}{\lambda}\) = A\(\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\) …. (1) Here, λ = wavelength of spectral line, A = constant and n = 3, 4, 5, ……… . The number of complete waves present in unit length is denoted by the reciprocal of wavelength, \(\frac{1}{\lambda}\) ; hence \(\frac{1}{\lambda}\) is called the wave number and sometimes expressed by the symbol f.

Substituting A = 1.09678 × 107m-1 in equation (1), the experimental values of the wavelength of the spectral lines can be found out. For example, for n = 3, λ = 6563Å; for n = 4, λ = 4861Å for n = 5, λ = 4341Å; for n = 6, λ = 4102Å

Moreover, substituting n = 7, 8, 9, ……. , different values of λ are obtained and these also belong to Balmer series. But these wave-lengths lie in the ultraviolet region, not in the visible region.

Balmer could arrange the spectral lines of hydrogen in a definite pattern, but could not determine relation (1) theoretically.

→ Lyman series: The relation denoting this series is, \(\frac{1}{\lambda}\) = A\(\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\); n = 2, 3, 4, ……

Using the same value of A, the values of λ obtained from this relation, resemble that of the lines obtained in the ultraviolet region of hydrogen spectrum. For example, if n = 2, then λ = 1216 Å, if n = 3, then λ = 1026 Å.

→ Paschen series: The relation denoting this series is, \(\frac{1}{\lambda}\) = A\(\left(\frac{1}{3^2}-\frac{1}{n^2}\right)\); n = 4, 5, 6,…….

From this relation, the wavelength of some spectral lines of the infrared region of hydrogen spectrum are obtained. For example, if n = 4, then λ = 18751 Å.

→ Brackett series and Pfund series: In the infrared region of hydrogen spectrum some more series are present except the Paschen series; these are Brackett series and Pfund series. But in this case, the brightness (or intensity) of the corresponding spectral lines is negligibly small.

The relation denoting the Brackett series is, \(\frac{1}{\lambda}\) = A\(\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\) ; n = 5, 6, 7, ….. and the relation denoting the Pfund series is, \(\frac{1}{\lambda}\) = A\(\left(\frac{1}{5^2}-\frac{1}{n^2}\right)\) ; n = 6, 7, 8, …..

Rydberg formula: Just after the discovery of Balmer series, Rydberg expressed the following general equation related with the series of spectrum. This is known as Rydberg formula. \(\frac{1}{\lambda}\) = \(\frac{R}{(m+a)^2}\) – \(\frac{R}{(n+b)^2}\) …. (2)

where, R is constant (Rydberg constant) for a particular element, a and b are the characteristic constants for a particular series, m is a whole number which is constant for a given series and n is a varying whole number whose different values indicate the different lines of the series. With the help of equation (2), the series of spectrum of most of the elements can be expressed accurately.

Rydberg constant: The discrete energy levels of hydrogen atom [equation (13) in section 1.2.3.] is, En = –\(\frac{R c h}{n^2}\) ….. (1) Here, R = \(\frac{m e^4}{8 \epsilon_0^2 c h^3}\) = constant …. (2) where, c = speed of light in vacuum = 3 × 108 m ᐧ s-1 and n = 1, 2, 3, ….. ∴ Ground state energy, E1 = -Rch Substituting the values of different constants In equation (2), we get, R ≈ 1.09737 × 107m-1.

Note that, in the analysis of Balmer series [section 1.2.4] of hydrogen spectrum, the value of the constant A obtained (1.09678 × 107m-1) is slightly less than the above value of R.

This difference becomes negligible if the mass of hydrogen nucleus is corrected. So, we can say that, the constant A is actually the Rydberg constant and equation (2) denotes its expression.

Calculation of Rydberg constant in CGS system: The expression for Rydberg constant in the CGS systçm can be obtained by substituting \(\frac{1}{4 \pi}\) in place of ϵ0 in equation (2). Using the CGS values of the constants, we get, R = \(\frac{2 \pi^2 m e^4}{c h^3}\) = 109737 cm-1

Wavelength of the emitted radiation: Let the electron in a hydrogen atom make a transition from a higher energy state to a lower energy state \(E_{n_f}\) [Fig.]. According to Bohr’s postulate, a photon will be emitted from the hydrogen atom due to this transition. If the frequency of this photon be f (wavelength, λ = \(\frac{c}{f}\)), then \(E_{n_i}\) – \(E_{n_f}\) = hf = \(\frac{h c}{\lambda}\) ….. (3) Substituting n = ni and n = nf respectively in equation (1), we get, \(\frac{R c h}{n_i^2}\) = –\(\frac{R c h}{n_i^2}\) and \(E_{n_f}\) = –\(\frac{R c h}{n_f^2}\) So, from the equation (3), we get \(\frac{h c}{\lambda}\) = –\(\frac{R c h}{n_i^2}\) + \(\frac{R c h}{n_f^2}\) = Rch\(\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\) or, \(\frac{1}{\lambda}\) = R\(\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\) …. (4)

Balmer series: If the electron in hydrogen atom jumps from any one of the energy states E3, E4, E5, ….., etc. to the energy state E2, then putting ni = 3, 4, 5, ….. and nf = 2 in equation (4), we get, \(\frac{1}{\lambda}\) = R\(\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\) ; n = 3, 4, 5, …… (5) This relation indicates the Balmer series of the atomic spectrum of hydrogen [section 1.2.4; equation (1)]

Other series: Similarly, substituting ni = 2, 3, 4, …… and nf = 1 in equation (4), we get the Lyman series: \(\frac{1}{\lambda}\) = R\(\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\) ; n = 2, 3, 4, …… (6) Again, substituting ni = 4, 5, 6, …… and nf = 3, we get Paschen series: \(\frac{1}{\lambda}\) = R\(\left(\frac{1}{3^2}-\frac{1}{n^2}\right)\) ; n = 4, 5, 6, ….. (7) Substituting ni = 5, 6, 7, ….. and nf = 4, we get Brackett series: \(\frac{1}{\lambda}\) = R\(\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\) ; n = 5, 6, 7, ……. (8) Substituting ni = 6, 7, 8, ….. and nf = 5, Pfund series is obtained: \(\frac{1}{\lambda}\) = R\(\left(\frac{1}{5^2}-\frac{1}{n^2}\right)\) ; n = 6, 7, 8, ……. (9) So the relations, shown by Balmer and other scientists for different wavelengths of the atomic spectrum of hydrogen, can be established accurately from Bohr’s theory. The basis of the suc-cess of Bohr’s theory lies in the accurate explanation of the hydrogen spectrum, although it has deviations from classical physics. Different series of the atomic spectrum of hydrogen are shown in Fig.

We have seen that if an electron jumps from a higher energy state to a lower one, energy equal to the difference between the states is emitted where ΔE = hc/λ. So the reverse process, if the atom absorbs a photon of wavelength λ, the electron will jump from the lower energy state to the higher energy state. Since the difference of energy states are fixed, the wavelength of the absorbed photon and that of the emitted photon are exactly equal.

Thus, if the light coming from a source (e.g., an incandes-cent tungsten lamp) pãsses through hydrogen gas, due to absorption, some dark lines are formed in its spectrum. These dark or black lines form an absorption spectrum [(a)]. The bright lines present in the emission spectrum obtained from hydrogen gas discharge tube, become dark in the continuous spectrum of other sources while absorbed by hydrogen [Fig.(b)].

Example 1. If the value of Rydberg constant of hydrogen is 109737 cm-1, determine the longest and the shortest wavelength of the Balmer series. [HS 04] Solution: If the wavelength of Balmer series be λ, then \(\frac{1}{\lambda}\) = R\(\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\) [n = 3, 4, 5, ……∞; R = Rydberg constant] Substituting the minimum value of n, i.e., n = 3, we get, \(\frac{1}{\lambda}\) = R\(\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\) = R × \(\frac{5}{36}\) or, λ = \(\frac{36}{5 R}\) = \(\frac{36}{5 \times 109737}\)cm = \(\frac{36 \times 10^8}{5 \times 109737}\)Å = 6651 Å This is the longest wavelength. Again, substituting the maximum value of n, i.e., n = ∞, we get, \(\frac{1}{\lambda}\) = R\(\left(\frac{1}{2^2}-\frac{1}{\infty}\right)\) = R × \(\frac{1}{4}\) or λ = \(\frac{4}{R}\) = \(\frac{4}{109737}\) cm = \(\frac{4 \times 10^8}{109737}\)Å = 3645Å This is the shortest wavelength.

Example 2. As a result of collision with an electron of 20 eV energy, a hydrogen atom Is excited to the higher energy state and the electron is scattered with a reduced velocity. Subsequently a photon with wave length 1216 A is emitted from the hydrogen atom. Determine the velocity of the electron after collision. Solution: Wavelength of the emitted photon, λ = 1216 Å = 1216 × 10-8 cm ∴ The amount of energy gained by the hydrogen atom from the electron, E2 – E1 = hf = \(\frac{h c}{\lambda}\) = \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{1216 \times 10^{-8}}\) =1.634 × 10-11 erg According to the questions, the initial kinetic energy of the electron = 20 eV = 20 × 1.6 × 10-12 erg = 3.2 × 10-11 erg. ∴ The remaining kinetic energy of the electron after its collision with hydrogen atom is, \(\frac{1}{2}\)mv2 = (3.2 – 1.634) × 10-11 = 1.566 × 10-11 erg. or, v = \(\sqrt{\frac{2 \times 1.566 \times 10^{-11}}{9.1 \times 10^{-28}}}\) = 1.855 × 108 ᐧ s-1.

Example 3. Light from a discharge tube containing hydrogen atoms is incident on the surface of a piece of sodium. The maximum kinetic energy of the photoelectrons emitted from sodium is 0.73 eV. The work function of sodium is 1.82 eV. Find (i) the energy of photons responsible for the photoelectric emission, (ii) the quantum numbers of the two orbits in the hydrogen atom involved in emission of photons and (iii) the change in angular momentum of the electron of hydrogen atom in the two orbits. Solution: i) According to the photoelectric equation, energy of the photon, hf = Emax + W0 = 0.73 + 1.82 = 2.55 eV

ii) Energy difference between the two orbits is 2.55 eV. Now in case of hydrogen atom, Energy in the ground state (n = 1), E1 = -13.6 eV Energy in n = 2 state, E2 = \(\frac{E_1}{2^2}\) = –\(\frac{13.6}{4}\) = -3.4 eV Energy in n = 3 state, E3 = \(\frac{E_1}{3^2}\) = –\(\frac{13.6}{9}\) = -1.51 eV Energy in n = 4 state, E4 = \(\frac{E_1}{4^2}\) = –\(\frac{13.6}{16}\) = -0.85 eV If calculations are done up to this state, it is seen that, E4 – E2 = 2.55 eV So, quantum numbers of the two orbits are 4 and 2.

iii) According to Bohr’s postulate, change in angular momen = 4 ᐧ \(\frac{h}{2 \pi}\) – 2 ᐧ \(\frac{h}{2 \pi}\) = \(\frac{h}{\pi}\) = \(\frac{6.625 \times 10^{-27}}{\pi}\) = 2.11 × 10-27 erg ᐧ s

Example 4. When ultraviolet light of wavelengths 800Å and 700Å are incident on the hydrogen atom at ground state, electrons are emitted with energies 1.8 eV and 4 eV, respectively. Determine the value of Planck’s constant. Solution: Let the ground state energy of hydrogen atom = -E0. Hence, the minimum amount of energy E0 is required to liberate its electron, i.e., the work function of hydrogen atom, W0 = E0. So, if the incident photon can provide E amount of kinetic energy to the electron, then hf = E + E0 or, \(\frac{h c}{\lambda}\) = E + E0 …… (1) In the first case, λ1 = 800 Å = 800 × 10-8cm, E1 = 1.8 eV= 1.8 × 1.6 × 10-12 erg In the second case, λ2 = 700 Å = 700 × 10-8cm, E2 = 4.0 eV = 4.0 × 1.6 × 10-12 erg From equation (1), we get,

Example 5. In absorbing 10.2 eV energy, the electron of a hydrogen atom jumps from its initial orbit to next higher energy orbit. As the electron returning to the former orbit, a photon is emitted. What is the wavelength of this photon? Solution: According to Bohr’s postulate, hν = Ei – Ef or, \(\frac{h c}{\lambda}\) = Ei – Ef or, λ = \(\frac{h c}{E_i-E_f}\) Difference between two energy levels, Ei – Ef = 10.2 eV = 10.2 × 1.6 × 10-12 erg ∴ λ = \(\frac{6.55 \times 10^{-27} \times 3 \times 10^{10}}{10.2 \times 1.6 \times 10^{-12}}\) = 1204 × 10-8 cm = 1204 Å