Question

Given the time-independent Schrodinger equation for a hydrogen atom to be

Answer 1

\[\frac{ -\hbar^{2} }{ 2m }\left[\frac{ d^2\psi }{ dr^2 }+\left(\frac{ 2 }{ r }\right)\left(\frac{ d\psi }{ dr }\right)\right]+\frac{ l(l+1) }{ r^2 } + U\psi = E\psi\]
and the potential energy function of the hydrogen atom to be\[U = \frac{ -ke^2 }{ r }\] where \[k = \frac{ 1 }{ 4\pi\epsilon _{0} }\] show that a valid solution for this form of the Schrodinger equation is\[\psi = \left( \frac{ 1 }{ \sqrt{4\pi} } \right)\left( \frac{ 2 }{ \sqrt{a _{0}^3} } \right)e ^{\frac{ -r }{ a _{0} }}\] for an electron in the n=1 energy state if \[E _{n}=-\frac{\left[ \frac{ e^4m _{e} }{ 2(4\pi\epsilon _{0})^2\hbar^2 } \right]}{n^2}\] and\[r _{0}=a _{0}=the 1st Bohr radius = 0.0529nm\]

Answer 2

So I know so far\[\frac{ d\psi }{ dr }=\left( \frac{ 1 }{ \sqrt{4\pi} } \right)\left( \frac{ 2 }{ \sqrt{a _{0}^3} } \right)\left( -\frac{ 1 }{ a _{0} } \right)e^\frac{ -r }{ a _{0} }\]\[\frac{ d^2\psi }{ dr^2 }=\left( \frac{ 1 }{ \sqrt{4\pi} } \right)\left( \frac{ 2 }{ \sqrt{a _{0}^3} } \right)\left( \frac{ 1 }{ a _{0}^2 } \right)e^\frac{ -r }{ a _{0} }\] which when plugged into the given Schrodinger equation becomes\[\frac{ -\hbar^2 }{ 2m }\left\{\left[ \left( \frac{ 1 }{ \sqrt{4\pi} } \right)\left( \frac{ 2 }{ \sqrt{a _{0}^3} } \right)\left( \frac{ 1 }{ a _{0}^2 } \right)e^\frac{ -r }{ a _{0} } \right]+\left[ \left( \frac{ 2 }{ r } \right)\left( \frac{ 1 }{ \sqrt{4\pi} } \right)\left( \frac{ 2 }{ \sqrt{a _{0}^3} } \right)\left( -\frac{ 1 }{ a _{0} } \right)e^\frac{ -r }{ a _{0} } \right] \right\}\]\[-\frac{ ke^2 }{ r }\psi=E\psi\] where the term \[\frac{ l(l+1) }{ r^2 }\] goes to zero because the electron is in the n=1 energy state.