EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 36, Problem 42P
To determine

The ground state hydrogen atom wave function is solution to which equation.

Expert Solution & Answer
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Answer to Problem 42P

The ground state hydrogen atom wave function is solution to Schrodinger’s equation in spherical coordinatesis 22mr2{r(r2ψr)+[1sinθθ(sinθ ψ θ)+1 sin2θ2ψϕ2]}+U(r)ψ=Eψ .

Explanation of Solution

Given:

The ground state hydrogen atom wave functionis ψ100=π1/2(Z/ a o)3/2eZr/ao .

Formula Used:

The expression for Schrodinger’s equation in spherical coordinates is given by,

  22mr2{r(r2ψr)+[1sinθθ(sinθ ψ θ)+1 sin2θ2ψϕ2]}+U(r)ψ=Eψ

The expression U(r) for is given by,

  U(r)=kZe2r

The expression for ao is given by,

  ao=2mke2

Calculation:

The ground state is having spherical symmetry. So,

  [1sinθθ(sinθψθ)+1sin2θ2ψϕ2]=0

The value of r(r2ψr) is calculated as,

  r(r2 ψ r)=r(r2 r( ( Z/ a o ) 3/2 e Zr/ a o π 1/2 ))=r(r2( ( Z/ a o ) 3/2 e Zr/ a o π 1/2 )× Z a o )= ( Z/ a o ) 3/2 π 1/2 ×Zao[2re Zr/ a o +r2e Zr/ a o ×Z a o]=re Zr/ a o π 1/2 ( Z a o )5/2[rZ a o2]

The Schrodinger’s equation in spherical coordinates is calculated as,

  22mr2{r( r 2 ψ r)+[1 sinθ θ( sinθ ψ θ )+1 sin 2 θ 2 ψ ϕ 2 ]}+U(r)ψ=Eψ22mr2{r e Zr/ a o π 1/2 ( Z a o )5/2[ rZ a o 2]+0}+( kZ e 2 r)( e Zr/ a o π 1/2 ( Z a o ) 3/2 )=Ee Zr/ a o π 1/2 ( Z a o )3/222mr(Z a o )[rZ a o2]+( kZ e 2 r)=E22mr(Z 2 mk e 2 )[rZ 2 mk e 2 2]+( kZ e 2 r)=E

Solve further as,

  22mr(Z 2 mk e 2 )[rZ 2 mk e 2 2]+( kZ e 2 r)=EE=Z2k2e4m22

The above expression for E is correct for the ground state energy.

Conclusion:

Therefore, the ground state hydrogen atom wave function ψ100=π1/2(Z/ a o)3/2eZr/ao is a solution to Schrodinger’s equation in spherical coordinates is 22mr2{r(r2ψr)+[1sinθθ(sinθ ψ θ)+1 sin2θ2ψϕ2]}+U(r)ψ=Eψ .

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