( ) into above PDE. ( ), wherec = 1

Transcription

1 xample of how to veify a Hydogen Solution The hydogen atom solution is pesented in section 7., equation 7.7, ψ nlml,θ,φ) R nl ) θ,φ ae shown in 7. and 7.. It is the solution of the patial diffeential equation PD), nlm l + nlm l θ + ψ nlml, see equation 7.3, " n 4πε 0 ψ 0 nlm l ), whee the exact fom of the lowest quantum numbes whee fom equation 7., n m e! 4πε 0 n 3.6eV, whee m µ is n effectively the mass of electon. See also equation 4.5 and 4.6. This can be veified by diect substitutionψ nlml,θ,φ ) into above PD. xample: Using Table 7. and 7., 4 ψ 3,,,θ,φ) R 3 )Y, θ,φ) 3/ 8 30 a exp 0 3a 0 4 ) C exp ψ 3,,,θ,φ 3a 0 sin θ exp iφ ), wheec 7/ π sin θ exp iφ ). 4 π, and 4πε 0! m 0.59 Å is the Boh adius discussed in equation 4.4. me We veify by diect substitution into time-independent Schodinge equation: 3,, + 3,, θ + ψ 3,,.! n 4πε 0 ψ 0 3,, Do the fist tem, 3,, C exp 3a 0 sin θ exp iφ ), ψ 3,, C sin θ exp iφ ) d d d d exp 3a, whee note that 0

2 d, This gives d 3,, C sin θ exp iφ ψ 3,, C sin θ exp iφ ) d ) d 3a 0 exp 3a, 0 3a 0 3a exp 3a 0. 3,, C sin θ exp iφ )exp 3a + 0. Using 0 ψ 3,,,θ,φ) C exp 3a 0 sin θ exp iφ ), 3,, ψ + 3,,. 0 Now we do the second tem, 3,, θ sinθ θ sinθ θ C exp 3a 0 sin θ exp iφ ), 3,, θ C exp d 3a exp iφ ) 0 sinθ dθ sinθ d dθ sin θ ), note how θ d dθ, 3,, θ C exp d 3a exp iφ ) 0 sinθ dθ sin θ cosθ 3,, θ C exp 3a exp iφ 0 3,, θ C exp 3a 0 sin θ exp iφ Usingψ 3,,,θ,φ ) C exp 3a 0 sin θ exp iφ ) 3,, θ ψ 4cos θ sin θ 3,, sin θ ), ) 4sinθ cos θ sin 3 θ ) sinθ ) 4cos θ sin θ ) sin θ ), and the identity cos θ sin θ, ψ 3,, 4 6sin θ ) sin θ,.

3 Now we do the thid tem, ψ 3,, C exp 3a 0 sin θ exp iφ ) ψ 3,, C exp d 3a 0 sin θ sin θ dφ exp iφ )), note ψ 3,, C exp 4 3a 0 sin θ exp iφ ) sin θ ψ 4 3,, sin θ. To summaize: 3,, ψ + 4 3,, ; 0 ) 3,, θ ψ 4 6sin θ 3,, sin θ ; sin θ Hence 3,, + 3., θ + ψ 3,, ψ 3,, ) + 4 6sin θ +ψ 3,, 0 sin θ +ψ 3,, 4 sin θ φ d dφ, ψ 3,, φ ψ 3,, 4 sin θ. ψ 3,, ψ ψ + 4 3,, 3,, a ,, + 3., θ + ψ 3,, ψ 3,, Substituting into the time-independent Schodinge equation fo n 3: 3,, + 3,, θ + ψ 3,,! 3 e 4πε 0 ψ 0 3,, ψ 3,, + 9! 3 4πε 0 ψ 0. 3,,

4 Using 4πε 0! and me n m e! 4πε 0 manipulation eveals 3 m e! 4πε 0 n 3 m 9 m/! 9 m 4πε 0! 4 me e! 4πε 0 ) m! m! 3! 8ma m e. Also note 0 9! 4πε 0!, whee some 9!, which gives 8m. Combining all the tems, ψ 3,, + 9! 3 4πε 0 ψ 0 3,, ψ 3,, + 9! 9a + 0 ψ 0 0 0, QD. Hence we have veified 3,, that ψ3,, is a solution of the time-independent Schodinge quation. Unsöld Theoem: Hydogen-like states in filled o half-filled subshell same quantum numbes n and l ) have spheical electon distibution, as quantified by the mathematical elation: m l l m l l ) ψ nlml f m l l l+ 4π. m l l, whee f) is a function of the adius only, o moe simply Consequence of Unsöld Theoem and the Anti-Symmety of lectonic States: Hydogen-like states in filled subshell same quantum numbes n and l ) contibute zeo angula momentum, L 0, S 0, J 0. xample of Unsöld theoem: Conside Manganese Mn) with Z 5 5 potons, 5 electons fo electical neutality). The electonic configuation iss s p 6 3s 3p 6 4s 3d 5, whee the unfilled subshell 3d 5 ) has five electons in a subshell with capacity of 0 l+) )+ m l ) 0 ), and the subshell is half filled and should obey Unsöld theoem: + 4π m l 5 4π. Fom table 7., Y, 4 π sin θ exp φi )

5 Y, * Y, Y, 4 π sin θ exp φi ) 4 students should be able to see thaty, 4 Y, * Y, Y, Y, Y, Y, Y, 4 sinθ cosθ exp iφ ) π π sin θ exp φi ) ) 4 sinθ cosθ exp iφ ) ) π sinθ cosθ exp iφ ) π sinθ cosθ exp iφ ) π Y, π 3cos θ ) Y,0 6π 3cos θ Now we ae eady to calculate m l m l π sin θ exp φi ) 3π sin4 θ. Similaly, π sin θ exp φi ) π sin θ exp φi ) 3π sin4 θ. sinθ cosθ exp iφ ) π 8π sin θ cos θ. sinθ cosθ exp iφ ) π 8π sin θ cos θ. ). Y, + Y, + Y,0 + Y, + Y,, m l 3π sin4 θ + 8π sin θ cos θ + 5 6π 3cos θ m l + 8π sin θ cos θ + 3π sin4 θ m l m l ) ), 6π sin4 θ + 4π sin θ cos θ + 5 6π 3cos θ m l 5 6π 3sin4 θ +sin θ cos θ + 3cos θ m l m l m l ), 5 6π 3sin4 θ +sin θ cos θ + 9cos 4 θ 6cos θ +), but sin 4 θ sin θ cos θ m l m l ) sin θ sin θ cos θ, which when substituted into st tem, ), 5 6π 3sin θ 3sin θ cos θ +sin θ cos θ + 9cos 4 θ 6cos θ + m l 5 6π 3sin θ + 9sin θ cos θ + 9cos 4 θ 6cos θ + m l ), and using,

6 cos 4 θ cos θ sin θ ) cos θ sin θ cos θ, which when substituted to 4 th tem, 5 6π 3sin θ + 9sin θ cos θ + 9cos θ 9sin θ cos θ 6cos θ +), m l m l m l 5 6π 3sin θ +3cos θ + m l Complex Numbes ) 5 4 π l+ 4π, fo l. QD Imaginay numbes: i i. Complex numbes: z x + iy, whee x, and y ae eal numbes. Fo example,..9i x., y.9. Real pat,re..9i ).9. Im..9i ).; Imaginay pat Complex Conjugate z * ): Fo a complex numbe z x + iy z * x iy. Some examples: z 0.6+i z * 0.6 i ; z 7.i z * 7+.i. Absolute Value): z zz * x + iy Pythagoas theoem. Fo example: z 0.6+i z * 0.6 i zz * ; z 4 i z * 4 + i zz * We can also wite z zz * x + y. Fo example, ) x iy) x + y, which is basically z 0.6+i z * 0.6 i z zz * 4.36 ; z 4 i z * 4 i z zz * 7. Pola Fom: Any complex numbe can be witten z x + iy exp iφ ), whee zz * x + y, and φ is the phase of the complex numbe. Anothe way to expess this is z x + iy exp iφ z * exp iφ ) cos φ)+ isin φ z zz * exp iφ ) exp iφ ) cosφ + isinφ). The complex conjugate is ) ) x iy. To veify ) zz * x + y. ) cosφ isinφ

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