6. Given real numbers $m, n$ satisfy $m-n=\sqrt{10}$, $m^{2}-3 n^{2}$ is a prime number. If the maximum value of $m^{2}-3 n^{2}$ is $s$ and the minimum value is $t$. Then the value of $s-t$ is ( ).
(A) 12
(B) 14
(C) 11
(D) 13
6.C.

Let $m^{2}-3 n^{2}=p$ (where $p$ is a prime number).
From $m-n=\sqrt{10} \Rightarrow(n+\sqrt{10})^{2}-3 n^{2}=p$
$$
\Rightarrow 2 n^{2}-2 \sqrt{10} n+(p-10)=0 \text {. }
$$

Since $n$ is a real number, then
$$
(-2 \sqrt{10})^{2}-4 \times 2(p-10) \geqslant 0 \text {. }
$$

Solving this, we get $p \leqslant 15$. Therefore, $p=2,3,5,7,11,13$.
Thus, $s=13, t=2, s-t=11$.