Let $L$ be an atomic ortholattice. We say that two elements $a$ and $b$ of $L$ are orthogonal if $a\leq b^\perp$. If $L$ is orthomodular then every element of $L$ can be written as a join of pairwise orthogonal atoms of $L$. Is the converse also true?
Yes there are. Any nonmodular finite orthomodular lattice is an example. A concrete example: take the Boolean algebras $2^2$,$2^3$ and identify their top and bottom elements. This is clearly an atomistic ortholattice, which is not modular, because it contains a pentagon as a sublattice. 


Examples are contained in a classical book about semimodular lattice theory: F. Maeda and S. Maeda "theory of symmetric lattices". See that book for details and refernces concerning what follows. Let $V$ be a infinite dimensional, separable, metrically incomplete preHilbert space (over real, complex or quaternionic scalars, or more generally any archimedean Baerordered $*$field). Let $L$ the collection of orthoclosed subspaces of $V$ (orthocosed means being the orthogonal of a subset of $V$, i.e. being the biorthogonal of itself). Ordered by inclusion, $L$ is a complete lattice (the lattice of
``closed'' set for the Galois connection given by orthogonality). $L$ is
orthocomplemented in the obvious way (the orthocomplement is the set of
orthogonal vectors). $L$ is atomistic (every element is join of atoms),
with atoms being the one dimensional linear subspaces of $V$ (the finite
dimensional linear subspaces are all orthoclosed, form a join dense modular
sublattice, and together with their orthocomplements form a modular
orthocomplemented lattice whose completion by cuts is canonically The babylonian method for completing the square (m\'ethode connue
\'egalment, dans dans la litt\'erature, sous le noms de "m\'ethode de
Jacobi" et "m\'ethode d'ortogonalization de Schmidt", `a moins qu'il
ne conviene plut\^ot de l'attribuer `a quelque savant russe, wrote
A. Weil in 1957) shows that each separable (and also each complete) To show that $L$ is not orthomodular precisely because it is metrically incomplete, now apply the theorem of Piron  Amemya / Araki  Morash  Gross / Keller  Wilbur  Holland. (Digression: One does not need the much deeper
and mathematically wonderful theorem of M. P. Soler; neither one does
need the even deeper, and unfortunately well forgotten by modern quantum
logicians, characterization by von Neumann of "continuous geometries
with transition probability". This characterization, even when applied
to its easier subcase, the type I finite case, provides an
axiomatization of Hilberian quantum logics of lenght at lest 4 which The above example $L$ has infinite lengh. I have no well known examples of finite lenght (I am familiar only with the semimodular case, which implies modularity in finite lenght with antiautomorphism). However, some related comments follow, ending in a candidate example. Note that the usual meaning of "atomistic" is: each element is join of atoms. Equivalently (see again Maeda  Maeda): atomic (each nonzero element contains an atom) and sectionally semicomplemented (if $x<y$ then $y$ contains a nonzero $z$ disjoint from $x$). Since "orthomodular" means "relatively orthocomplemented" (if $x<y$ then a $z$ orthogonal to $x$ exists whose join with $x$ is $y$), one sees that "atomic orthomodular" imples atomistic. Besides, for complete orthocomplemented atomic lattices, one sees that orthomodular is equivalent to "each set of orthogonal atoms contained in a element $x$ expands to one such set with join $x$", and is also equivalent to "each maximal orthogonal set of atoms contained in a element $x$ has join $x$". These equivalent conditions for orthomodularity are strictly stronger than "each element is a orthogonal join of atoms" as shown by incomplete separable preHilbert spaces. On the other hand, atomistic is strictly weaker than "each element is a join of orthogonal atoms". Infact, the following method produces finite, atomistic orthocomplemented lattices which are not orthomodular because some elements have no orthogonal basis of atoms. Take any finite orthoalgebra $A$ which is not a lattice (examples are frequently given in literature, using Greechie diagrams). Let $L$ be its
cut (normal, Dedekind  McNeille) completion. $L$ is complete, even
finite, othocomplemented lattice (as any completion of a orthocomplemented, even finite, orthoposet $A$). $L$ is atomistic since Two final, and formally offtopic (but hopefully intuitively helpful) remarks (before the appendix with the candidate example). First, one can easy also give atomistic orthocomplemented examples
where there are many elements besides atoms and the top / bottom Second remark. To any orthocomplemented lattice $L$ one can associate at
least two orthomodular posets (which coincide with $L$ iff $L$ is
orthomodular). First, consider the same set $L$, but this time with a stronger ordering: $x$ has a orthocomplement in $y$ (as computed in $L$;
one could say: $x$ splits $y$). One obtains a $L'$ If there is a finite lenght orthocomplementd lattice $L$, not orthomodular but with orthogonal bases of atoms for each element, then comparing $L$ with $L'$ and $L''$ should be instructive, and so should be comparing with the set of elements which have a orthogonal basis of atoms (note that atoms give a Foulis  Randall manual, with a associated orthoalgebra consisting exactly of such elements). [Form Henstock (his general theory of integration, in division spaces), a Appendix: Tentative candidate finite counterexample: first, take the horizontal sum of two copies of the boolean algebra with two atoms $2^3$ (or more
generally two finite orthomodular lattices $A$, with an atom $a$ whose
orthocomplement $a'$ is not an atom, and $B$, with an atom $b$ whose
orthocomplement $b'$ is not an atom). Then, to this orthomodular 

