Lecture3 Atoms Electrons and Bonding

x a sin kx n k n 1 2 3 l 2men n n is

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Unformatted text preview: equations ∂ 2 Ψ ( x ) 2m + [ E − V ( x )] Ψ( x ) = 0 2 2 ∂x ∂ E Φ( t ) − Φ( t ) = 0 ∂t i EE 332 Spring 2013 Simple Potential Well Problem Electron is trapped in a potential well with V(x) = 0, 0 < x < L and V(x) = ∞, x = 0, L Inside the well, V(x) = 0 ∂ 2 Ψ ( x ) 2m + E Ψ( x ) = 0 2 2 ∂x Possible solutions to the wave equation are sin(kx) and cos(kx) where k = 2mE / Boundary condition ψ(x=0, L) = 0. Ψ( x ) = A sin( kx ) nπ k= , n = 1, 2, 3, ... L 2mEn nπ n is quantum number = L ψ and E n n 222 nπ En = 2mL2 EE 332 Spring 2013 Quantum Number and Orbitals Quantum Number and Orbitals −1 q V ( r ,θ ,φ ) = V ( r ) = −( 4πε0 ) 2 r Ψ( r ,θ ,φ ) = R( r )Θ( θ )Φ( φ ) Ψ nlm ( r ,θ ,φ ) = Rn ( r )Θl ( θ )Φm ( φ ) Pauli exclusion principle : no two electrons can have the same set of quantum number n, l, m, s EE 332 Spring 2013 Quantum Number QQuantumNumber and Orbital of Elements uantum Number and Orbital of Elements and Orbital of Elements EE 33...
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