What is the kinetic energy of the hydrogen nucleus? We now that from this equation:
$$\hat {H} (r , \theta , \varphi ) \psi (r , \theta , \varphi ) = E \psi ( r , \theta , \varphi) $$
we can derive the energy of a system. Total electron energy for hydrogen atom is:
$E_n = -\dfrac {m_e e^4}{8\epsilon_0^2 h^2 n^2}$ using central force potential.
But what is the kynetic energy of nucleus?
 A: There are three terms in the Schrodinger Hamiltonian $\hat H$ for hydrogen: the kinetic energy of the electron ($\hat{K}_e$), the kinetic energy of the proton ($\hat{K}_p$), and the electrostatic potential energy between them ($\hat{U}$). After reducing the equation for the two-particle wavefunction $\Psi(\mathbf{r}_1,\mathbf{r}_2)$ to a relative-separation wavefunction $\Psi(\mathbf{r})$, it has the form
$$\hat{H}\Psi=(\hat{K}_e+\hat{K}_p+\hat{U})\Psi=\left(-\frac{\hbar^2}{2m_e}\nabla^2-\frac{\hbar^2}{2m_p}\nabla^2-\frac{e^2}{4\pi\epsilon_0r}\right)\Psi=E\Psi.$$
The ground state is
$$\Psi_1=\frac{e^{-r/a_0}}{(\pi a_0^3)^{1/2}}$$
where
$$a_0=\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}$$
is the Bohr radius. Here
$$\mu=\frac{m_pm_e}{m_p+m_e}$$
is the reduced mass of the electron-proton system. The ground state has energy
$$E_1=-\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2}\approx-13.6\text{ eV}.$$
In this ground state, the expectation value of the proton kinetic energy operator is easily calculated to be
$$\begin{align}
\langle{K_p}\rangle&=\left\langle\Psi_1\left|-\frac{\hbar^2}{2m_p}\nabla^2\right|\Psi_1\right\rangle\\
&=\frac{\hbar^2}{2m_pa_0^2}=\frac{\mu^2e^4}{2m_p(4\pi\epsilon_0)^2\hbar^2}=-\frac{\mu}{m_p}E_1=-\frac{m_e}{m_p+m_e}E_1\\
&\approx 7.40\text{ meV}.
\end{align}$$
(Note: meV is milli-electron-volts, not mega-electron-volts!)
For other energy states, rather than computing the expectation value of the proton kinetic energy operator for a general hydrogen orbital $\Psi_{nlm}$, it is easier to use an argument based on the quantum virial theorem which in the case of a Coulombic potential says that
$$2\langle\hat{K}_e+\hat{K}_p\rangle=-\langle\hat{U}\rangle.$$
Combining this with the well-known orbital energies
$$\langle\hat{H}\rangle=\langle\hat{K}_e+\hat{K}_p+\hat{U}\rangle=E_n=\frac{E_1}{n^2}$$
one finds that
$$\langle\hat{K}_e+\hat{K}_p\rangle=-E_n=-\frac{E_1}{n^2}.$$
From the form of the kinetic energy operators, it is clear that the relative contributions of the electron and proton are in the ratio $1/m_e$ to $1/m_p$, so for any orbital
$$\langle\hat{K}_p\rangle=-\frac{m_e}{m_p+m_e}E_n\approx\frac{7.40\text{ meV}}{n^2}.$$
Note: Since 2-1/2 years have passed without anyone voting to close this question as homework-like, I felt it was OK to post a complete answer.
A: $E_n$ is the total energy of the electron-nucleus system because it contains a potential energy term which accounts for the electrostatic interaction between the electron and the nucleus.  
The nucleus is much heavier than the electron so it's considered as staionary, hence it's kinetic energy is zero. But this is just an approximation (a good one though), both the electron and the nucleus rotate about their common center of mass, and one can easily attack this problem by using the center of mass coordinates and reduce this two body problem to a one body problem and introduce the reduced mass of the system.  
The mathematical analysis is strightforward and you can find it in many references. you'll find a similar formula of the total energy:
$$E_n = -\dfrac {\mu e^4}{8\epsilon_0^2 h^2 n^2}$$
where $\mu$ is the reduced mass which is given by $\mu={m_em_p \over m_e+m_p}$ with $m_e$ and $m_p$ are the mass of the electron and the proton (the nucleus). You can see clearly that $\mu \simeq\ m_e$ due to the fact that $m_p$ is much greater than $m_e$. Thus your formula is a good approximation.  
One of the benefits of considering the motion of the nucleus is the theoretical  prediction of the existing of neutrons! By analysing the spectra of the hydrogen gaz which may contain both hydrogen and deuterium, we observe a slight shift in the observed sepctral lines of the ordinary hydrogen due the presence of deuterium, the Ryidberg formula of these two atoms allow us to find the ratio:
$${m_H \over m_D}\simeq\ 0.5$$
with $m_H$ is the mass of the hydrogen nucleus (the proton) and $m_D$ is the mass of the deuterium nucleus (proton+neutron).  
In other words, the atomic nuclei don't contain only protons, but also a neutral particles with approximately the same mass as protons called neutrons which were discoverd experimentally many years later by Chadwick.
