So isn't this a contradiction in the Bohmian picture of the atom?
In the Bohmian view, what would be kinetic energy can be 'stored' in the guiding wave. Here's an extensive excerpt from Reflections on the deBroglie–Bohm Quantum Potential
Consider the simple case of a cubical box with no classical potential
inside and infinite potential outside. In other words, the particle
cannot escape while the box remains intact. If an initially free
(spinless) particle is trapped in such a box of side length $L$, then
its wavefunction would take on the stationary waveform:
$$\psi =
(2L)^{3/2}|\sin(n_1\pi\,x/L)\sin(n_2\pi\,y/L)\sin(n_3\pi\,z/L)|e^{iE_nt/\hbar}
= Re^{iS/\hbar}$$
with total energy $E=(n^2_1 + n^2_2 + n^2_3)(\pi^2\hbar^2/2mL^2)$;
where $n_1,\,n_2,\,n_3$ are positive integers. In Orthodox Quantum
Theory, the particle is assumed to have kinetic energy only and to be
bouncing back and forth between the walls of the box. In
deBroglie–Bohm Theory, the value of the quantum potential is given
by: $Q = -(\hbar^2/2m)(\nabla^2R)/R = (n^2_1 + n^2_2 +
n^2_3)(\pi^2\hbar^2/2mL^2)$: This is the same magnitude as the
particle’s kinetic energy is assumed to have in Orthodox Quantum
Theory. However, since $S = -Et,\,\nabla S = 0$; i.e., the particle
has zero momentum and therefore zero kinetic energy. All the energy of
the system is potential with the kinetic energy of the quantum
particle having become stored in the wave field (Bohm1952, 184;
Riggs1999, 3072). What’s more, this energy will be returned to the
particle if the wave field’s stationary state is disturbed, e.g., if
any side of the box is removed. Surprisingly, this explanation was
originally suggested by David Bohm when he wrote:
... the kinetic energy of the particle will come from the $\psi$
field, which is able to store up even macroscopic orders of energy
when its wave-length is small (1953, 14, italics mine).