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MOLECULAR TRANSPORT SELF STUDY- 1
D8A.2 List and describe the significance of the quantum numbers needed to specify the internal state of a hydrogenic atom. Answer: Quantum numbers are a way to describe the discrete states that a particle can hold. There are four different quantum numbers needed to specify the state of an electron in hydrogen atom. These can the be used for solutions to the schrödinger equation. The three quantum numbers describe the size, shape and orientation of the atomica orbitals in space. There is an additional quantum number which does not follow from the schrödinger equation but is introduced to account for electron spin.
- The principal quantum number n gives the total energy of the orbital. n=1,2,3,…
- The orbital quantum number l gives the magnitude of the angular momentum(l can take an integer values from 0 to (n-1)). l=0,1,2,…,n- 1
- The magnetic quantum number ml gives the direction of the electron’s angular momentum(ml can take an integer values from - l to +l). ml=-l,-l+1,…,0,…,l-1,l or 2l+1 values
- The spin quantum number ms describes the angular momentum of an electron. ms=+1/2,-1/ E8A.2(a) The wavefunction for the ground state of a hydrogen atom is. Evaluate the normalization constant N. Answer: probability amplitude probability density
the normalization constant E8A.2(b) The wavefunction for the 2s orbital of a hydrogen atom is. Evaluate the normalization constant N. Answer: probability amplitude probability density
Substitute na/2z for r, Substitute 3 for n and 1 for z, wave function is normalised to 1. the wavefunctions are mutually orthogonal. b) Nodels are points in the orbital region where the probability of finding an electron is zero. It can be point lise or plane like. Nodal planes are perpendicular to the axis of symmetry. N=n-l-1 (the number of nodels) for 3s orbital: n=3 l= N=3- 0 - 1= for 3px orbital: n=3 l=1 and symmetrix axis is x N=3- 1 - 1=1 yz plane
for 3dxy orbital: n=3 l= N=3- 2 - 1= D8B.2 Outline the electron configurations of many-electron atoms in terms of their location in the periodic table. Answer: In a periodic table, various atoms are arranged in the increasing order of atomic number. A periodic table consists up of seven periods(horizontal rows of elements) and eighteen groups(vertical column of element). The different atoms are also grouped into four different blocks, namely s-block, p-block,d-block, and f-block based upon the orbital in which the differentiating electron is accommodated. In an s-block element, the differentiating electron is accommodated to s- orbital, but for a p-orbital, the differentiating electron is accommodated to a p-orbital for a d-block element, and for an f-block element, the differentiating electron is accommodated to a d-orbital and f-orbital respectively. We can easily predict the electronic configuration of an atom simply by observing their position within the periodic table. As we know there is a general formula for calculating the electronic configuration of each block of elements and hence it can be used to identify the electronic configuration of an atom simply by observing whether the element is in s or p or d or f-block. The general electronic configuration of the different blocks of elements are, (1) for an s- block element the general electronic configuration id ns^1 -^2. (2) for a p- block element the general electronic configuration id ns^2 np^1 -^6. (3) for a d- block element the general electronic configuration id (n-1)d^1 -^10 ns^1 -^2. (4) for an f- block element the general electronic configuration id (n−2)f(0−^14 )(n−1)d(0−1)ns^2. By using this general formula and by observing the position of an atom in a periodic table we can easily predict the electronic configuration of different multielectron atoms. E8B.4(a) Write the electronic configuration of the Ni2+ ion. E8B.4(b) Write the electronic configuration of the O2− ion. Answer: a) 1s^2 2s^2 2p^6 3s^2 3p^6 3p^8 b) 1s^2 2s^2 2p^6
For 1 S 0 , values of S, L and J are 0, 0 and 0. So values would be: ∆S=0-1=- 1 ∆L=0-1=- 1 ∆J=0-2=- 2 Transition is forbidden between two terms. iii) Transition: 3 F 4 → 3 D 3 For symbol 3 F 4 value of multiplicity is 3: (2S+1)=3 S= For symbol 3 D 3 value of multiplicity is 3: (2S+1)=3 S= For 3 F 4 , values of S, L and J are 1, 3 and 4. For 3 D 3 , values of S, L and J are 1, 2 and 3. So values would be: ∆S=1-1= ∆L=2-3=- 1 ∆J=3-4=- 1 Transition is allowed between two terms. b) i) Transition: 2 P3/2 → 2 S1/ For 2 P3/2, values of S, L and J are 1/2, 1 and 3/2. For 2 S1/2 , values of S, L and J are 1/2, 0 and 1/2. So values would be: ∆S=1/2- 1 /2= ∆L= 1 - 0 = ∆J=3/2-1/2= Transition between 2 P3/2 → 2 S1/2 is allowed. ii) Transition: 3 P 0 → 3 S 1 For 3 P 0 , values of S, L and J are 1, 1 and 0. For 3 S 1 , values of S, L and J are 1, 0 and 1. So values would be:
∆S=1-1=
∆L=1-0=
∆J=0- 1 =- 1
Transition between 3 P 0 → 3 S 1 is allowed. iii) Transition: 3 D 3 → 1 P 1 For 3 D 3 , values of S, L and J are 1, 2 and 3. For 1 P 1 , values of S, L and J are 0, 1 and 1. So values would be: ∆S=1-0= ∆L=2-1= ∆J=3-1= Transition between 3 D 3 → 1 P 1 is not allowed.
