The legs of a right triangle are in the ratio $19: 3$, then the length of one leg is ( ).
(A) $\frac{1}{2}\left(1993^{2}+1\right)$
(B) $\frac{1}{2}\left(1993^{2}-1\right)$
(C) has more than one but a finite number of values
(D) has an infinite number of values
7. (B).

Let the other leg be $a$, and the hypotenuse be $c$, then we have
$$
1993^{2}=c^{2}-a^{2}=(c+a)(c-a) .
$$

Since 1993 is a prime number, we have
$$
\left\{\begin{array}{l}
c+a=1993^{2}, \\
c-a=1 .
\end{array}\right.
$$

Thus, $a=\frac{1}{2}\left(1993^{2}-1\right)$.