State Bohr’s postulate to define stable orbits in hydrogen atom. How does de Broglie’s hypothesis explain the stability of these orbits ?

Bohr's postulate to define stable orbits in hydrogen atoms states that an electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of n h / 2 π , where h is Planck’s constant and n is the principal quantum number of orbit. Thus, an electron can revolve only in certain orbits of definite radii. These are called stable orbits. An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit corresponds, to a certain energy level. De Broglie hypothesis explanation of the stability of orbits: According to de Broglie, an electron orbiting around the nucleus is associated with a stationary wave. Electron wave is a circular standing wave. Since destructive interference will occur if a standing wave does not close upon itself, only those de Broglie waves exist for which the circumference of the circular orbit contains a whole number of wavelengths i.e., the circumference of n t h stable orbit for an electron can contain exactly n wavelength of de-Broglie. ⇒ n λ = 2 πr n n h p = 2 πr n n h m v = 2 πr n m v r n = n h 2 π ⇒ L = n h 2 π This was the quantum condition proposed by Bohr.

Calculate the orbital period of the electron in the first excited state of the hydrogen atom.

The radius of the n th orbit of the hydrogen and hydrogen like atoms is given by r n = 0 . 53 n 2 Z A ° where Z is atomic number. The velocity of the electron in n th orbit of hydrogen and hydrogen like atoms is given by v n = 2 . 2 × 10 6 × Z n m s - 1 The orbital period of the electron is given by T = 2 πr n v n Substituting the values from above, we get, T = 2 π × n × 0 . 53 × n 2 × 10 - 10 2 . 2 × 10 6 × Z × Z = 1 . 06 πn 3 × 10 - 10 2 . 2 × 10 6 × Z 2 For hydrogen atom, Z = 1 and for first excited state n = 2 . Put these values in above equation, we get T = 1 . 06 π 2 3 × 10 - 10 2 . 2 × 10 6 = 12 . 1 × 10 - 16 s . So, the orbital period of the electron in the first excited state is 1 . 21 × 10 - 15 s .

Explain the different types of spectral series.

When an electron in an excited state falls back to a lower energy state, it emits photon of particular energy which is equal to the difference in energies of two states. Rydberg formula gives a relation between energy difference between various energy levels and the wavelength of the photons. ν = R Z 2 1 n 1 2 - 1 n 2 2 where R is the Rydberg constant, Z is the atomic number and n 1 and n 2 are the principal quantum numbers of the orbit. The spectral lines are formed as a result of electronic transition from one energy level to another. For a hydrogen atom, these lines are divided into five series: (i) Lyman series: Lyman series is obtained when an electron jumps to the first orbit from any other outer orbit. The wavelengths of Lyman series lies in the ultraviolet region of the spectrum. (ii) Balmer series: Balmer series is obtained when an electron jumps to the second orbit from any other outer orbit. The wavelengths of Balmer series lies in the visible region of the spectrum. (iii) Paschen series: Paschen series is obtained when an electron jumps to the third orbit from any other outer orbit. The wavelengths of Paschen series lies in the infrared region of the spectrum. (iv) Brackett series: Brackett series is obtained when an electron jumps to the fourth orbit from any other outer orbit. The wavelengths of Brackett series lies in the infrared region of the spectrum. (v) Pfund series: This series is obtained when an electron jumps to the fifth orbit from any other outer orbit. The wavelengths of Pfund series lies in the infrared region of the spectrum.

E M B I B E

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 12 - Atoms>Exercise>Q 29

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12th Meghalaya Board

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The ratio of wavelengths of the last line of the Balmer series and the last line of the Lyman series is_____.

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Important Questions on Atoms

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12th Meghalaya Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 12 - Atoms>Exercise>Q 30

The following statements are given about the Hydrogen atom.

A. The wavelengths of the spectral lines of the Lyman series are greater than the wavelength of the second spectral line of the Balmer series.

B. The orbits correspond to circular standing waves in which the circumference of the orbit equals a whole number of wavelengths.