Radial Schrödinger equation I found a problem that says:

Show by direct substitution that $R_{10}$ is a solution of Schrödinger's radial equation.

AFAIK Schrödinger's radial equation is
$\frac{-\hbar^2}{2m}[\frac{d^2}{dr^2}+\frac{2}{r}\frac{d}{dr}-\frac{l(l+1)}{r^2}]R+E_p(r)R=ER$
and $R_{10}$ in this case is:
$2(\frac{Z}{a_0})^{3/2}e^{-\rho/2}$
where $\rho$ is
$\rho = \frac{2Zr}{na_0}$
and $E_p(r)$ is:
$E_p(r) = -\frac{Ze^2}{4\pi\epsilon_0r}$
The thing is I don't understand very well what $E$ is in this case and (I think) I need it to solve the problem. It's the total energy but in the book (Alonso y Finn) they never talk about the kinetic energy. I suppose it depends on $r$, $n$ and $l$ because once I do the substitution on the left side of the equation and I factorize it I get $R_{10}$ multiplied by a term that's full of those variables.
 A: $E$ is the total energy. Since the Hamiltonian $\hat{H}$ is 
$\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V$
in a time-independent Schrodinger equation
$\hat{H}\psi = E\psi$
in a spherical coordinates, you can write the wave function as:
$\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$
The Laplace operator $\nabla^2$ in the spherical coordinates is:
$\nabla^2 = \frac{1}{r}\frac{\partial}{\partial r}\left( r\frac{\partial}{\partial r}\right) + \frac{1}{r^2sin\theta}\frac{\partial}{\partial \theta}\left( sin\theta\frac{\partial}{\partial\theta} \right)+\frac{1}{r^2sin^2\theta}\frac{\partial^2}{\partial\phi^2}$
Put the $\nabla^2$ into the Schrodinger equation and separate the variables, you can get the radial equation in your question.
$-\frac{\hbar^2}{2m}\left[\frac{d^2}{dr^2} +\frac{2}{r}\frac{d}{dr}-\frac{l(l+1)}{r^2}\right]R + E_p(r)R = ER$
This equation is the Legendre's differential equation, where $l$ is the eigen value of the colatitude equation (equation of variable $\theta$) and $l$ can only be non-negative integers. This $l$ is also called the orbital quantum number. By solving this differential equation, you can solve the radial part $R_{n,l}(r)$ and the principle quantum number $n$, which yields the the state energy $E_n$.
Since the $-\frac{\hbar^2}{2m}\nabla^2$ is the kinetic energy part in Hamiltonian, so I think this solves your second question.  
