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The normalized radial probability density givesConsequently the expectation value of the potential energy isHere the last equality uses $Z{e}^2/(4\pi {ϵ}_0)={\hslash }^2\beta /m_e$. Since this is an energy eigenstate, $E=⟨T⟩+⟨V⟩$, and so $\boxed{⟨T⟩=-E}$. The kinetic energy is positive because $E<0$, in agreement with the virial theorem for a Coulomb potential.

The probability of finding the electron in a spherical shell is $P(r)dr=4\pi r^2|\psi (r){|}^{2}dr$, soThis radial probability density, rather than the density per unit volume, determines the most probable radius. Its expectation value isFor its interior maximum, differentiate $\log P=\text{constant}+2\log r-2\beta r$. The stationary condition $2/r-2\beta =0$ givesThe density per unit volume $|\psi {|}^2$ is instead largest at the nucleus; including the spherical-shell volume factor is essential to the requested radial mode.

The radial probability density $P(r)$ is defined so that $P(r)dr$ is the probability of finding a particle between radii $r$ and $r+dr$. The spherical volume element gives $P(r)=r^2\int |\psi (r,\Omega ){|}^{2}d\Omega$. For a spherically symmetric wavefunction, $P(r)=4\pi r^2|\psi (r){|}^2$; for $\psi =R(r)Y_l^m(\Omega )$ with normalized spherical harmonics, $P(r)=r^2|R(r){|}^2$. Its maximum need not coincide with the maximum of the density per unit volume: a hydrogenic $1s$ state has its largest spatial density at the origin but its most probable radius at $a_0/Z$.