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If Bohr’s quantisation postulate (angula...

If Bohr’s quantisation postulate (angular momentum = `nh//2pi` ) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?

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Bohr's quantisation postulate is in terms of Plank's constant (h), But angular momenta associated with planetary motion are `= 10^(70)` h (for earth): In terms of Bohr's quantisation posulate this will corresponq to `n = 10^(7).` For such large values of n the differences in successive energies and angular momenta of the quantised levels are so small, that the levels can be considered as continuous and not discrete.
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It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition must compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) he permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The maximum orbital angular momentum of an electron with n= 5 is

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition must compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) the permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The orbital angular momentum of an electron in p-orbital makes an angle of 45^@ from Z-axis. Hence Z-component of orbital angular momentum of election is :

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition mush compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) the permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The spin-only magnetic moment of free ion is sqrt(8) B.M. The spin angular momentum of electron will be

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An orbital is designated by certain values of first three quantum numbers (n, l and m) and according to Pauli.s exclusion principle, no two electrons in a atom can have all the for quantum numbers equal. N, l and m denote size, shape and orientation of the orbital. The permissible values of n are 1,2,3.... prop while that of 1 are all possible integral values from 0 to n-n. Orbitals with same values of n and 1 but different values of m (where m can have any integral values from 1 to +1 including zero) are of equal energy and are called degenerate orbitals. However degeneracy is destroyed in homogeneous external magnetic field due to different extent of interaction between the applied field and internal electronic magnet of different orbitals differing in orientations. In octahedral magnetic field external magnetic field as oriented along axes while in tetrahedral field the applied field actas more in between the axes than that on the axes themselves. For 1=0, 1,2,3,...., the states (called sub-shells) are denoted by the symbol s,p,d,f.....respectively. After f, the subshells are denoted by letters alphabetically 1 determines orbital angular motion (L) of electron as L = sqrt(l(l+1))(h)/(2pi) ON the other hand, m determines Z-component of orbital angular momentum as L_(Z) = m((h)/(2pi)) Hund.s rule states that in degenerate orbitals electrons do not pair up unless and until each each orbitals has got an electron with parallesl spins Besides orbital motion,an electron also posses spin-motion. Spin may be clockwise and anticloskwise. Both these spin motions are called two spins states of electrons characterized by spin Q.N (s) : s = +(1)/(2) and = -(1)/(2) respectively The sum of spin Q.N. of all the electrons is called total spin(s) and 2s+1 is called spin multiplicity of the configuration as a whole. The spin angular momentum of an electron is written as L_(s) = sqrt(s(s+1))(h)/(2pi) According to Hund.s rule, the distribution of electron within the various orbitals of a given sub-shell is one which is associated with

The behaviour of an electron in an atom is described mathematically by a wave function, or orbital. Spin of the electron produce angular momentum equal to S= sqrt(s(s+1)) (h)/(2pi) where S= +1/2 . Total spin of an atom =+ n/2 or h/2 Where n is the number of unpaired electron. The substance which contain species with unpaired electrons in their orbitals behave as paramagnetic substances. The paramagnetism is expressed in terms of magnetic moment the magnetic moment of an atom mu_s sqrt(s(s+1)) (eh)/(2pi mc) = sqrt((n/2)(n/2+1)) (eh)/(2pi mc) s = n/2 implies mu_s = sqrt(n(n+2)) B.M n = number of unpaired electrons 1 B.M. (Bohr magneton) = (eh)/(4pi mc) If magnetic moment is zero the substances is di-magnetic. If an ion of ""_(25)Mn has a magnetic of 3.873 B.M. Then Mn is in which state.

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{:("LIST - 1","LIST-2"),("A) Bohr's atomic","1) Fine spectrum of"),("model","Hydrogen"),("B) de-Broglie's","2) Atomic orbital"),("concept",),("C) Sommerfield","3) Dual nature of any"),("atomic model ","particle in motion"),("D) Schrodinger","4) Quantisation of"),("wave equation","angular momentum"):} The correct match is

VIKRAM PUBLICATION ( ANDHRA PUBLICATION)-ATOMS-ADDITIONAL EXERCISES
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  2. A hydrogen atom initially in the ground level absorbs a photon, which ...

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