= Solution
The normalized <radial probability density> gives
$$
\left\langle\frac1r\right\rangle=4\beta^3\int_0^\infty re^{-2\beta r}\,dr=\beta.
$$
Consequently the <expectation value> of the <potential energy> is
$$
\langle V\rangle=-\frac{Ze^2\beta}{4\pi\epsilon_0}=-\frac{\hbar^2\beta^2}{m_e}=\boxed{2E}.
$$
Here the last equality uses $Ze^2/(4\pi\epsilon_0)=\hbar^2\beta/m_e$. Since this is an <energy eigenstate>, $E=\langle T\rangle+\langle V\rangle$, and so $\boxed{\langle T\rangle=-E}$. The <kinetic energy> is positive because $E<0$, in agreement with the <virial theorem> for a <Coulomb potential>.
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