Question

The radial part of the Schrödinger equation for a system of two particles, of charges +e and -e, interacting through a Coulombic potential is: \frac{-\hbar^2}{2\mu r^2} \frac{d}{dr} \left( r^2 \frac{dR(r)}{dr} \right) - \frac{e^2 R(r)}{4\pi \epsilon_0 r} + \frac{\hbar^2}{2\mu r^2} l(l+1) R(r) = E R(r). (a) Show that the function: $R(r) = \frac{1}{4\sqrt{2\pi}} \left( \frac{1}{a_0} \right)^{3/2} \left( 2 - \frac{r}{a_0} \right) \exp\left( -\frac{r}{2a_0} \right)$, with $l = 0$, is a solution of the radial Schrödinger equation with the energy eigenvalue $E = \frac{-\mu e^4}{8\hbar^2 (4\pi \epsilon_0)^2}$. Take $a_0 = \frac{\hbar^2 (4\pi \epsilon_0)}{\mu e^2}$.

          The radial part of the Schrödinger equation for a system of two particles, of
charges +e and -e, interacting through a Coulombic potential is:
\frac{-\hbar^2}{2\mu r^2} \frac{d}{dr} \left( r^2 \frac{dR(r)}{dr} \right) - \frac{e^2 R(r)}{4\pi \epsilon_0 r} + \frac{\hbar^2}{2\mu r^2} l(l+1) R(r) = E R(r).
(a) Show that the function:
$R(r) = \frac{1}{4\sqrt{2\pi}} \left( \frac{1}{a_0} \right)^{3/2} \left( 2 - \frac{r}{a_0} \right) \exp\left( -\frac{r}{2a_0} \right)$, 
with $l = 0$, is a solution of the radial Schrödinger equation with the energy
eigenvalue $E = \frac{-\mu e^4}{8\hbar^2 (4\pi \epsilon_0)^2}$. Take $a_0 = \frac{\hbar^2 (4\pi \epsilon_0)}{\mu e^2}$.

The radial part of the Schrödinger equation for a system of two particles, of
charges +e and -e, interacting through a Coulombic potential is:
(-ħ^2)/(2μr^2) (d)/(dr) ( r^2 (dR(r))/(dr) ) - (e^2 R(r))/(4πϵ0 r) + (ħ^2)/(2μr^2) l(l+1) R(r) = E R(r).
(a) Show that the function:
R(r) = (1)/(4√(2π))( (1)/(a0))^3/2( 2 - (r)/(a0)) ( -(r)/(2a0)),
with l = 0, is a solution of the radial Schrödinger equation with the energy
eigenvalue E = (-μ e^4)/(8ħ^2 (4πϵ0)^2). Take a0 = (ħ^2 (4πϵ0))/(μ e^2).

Added by Laura J.

Question

Please give Ace some feedback