7. Let $A_{1}$ and $A_{2}$ be the left and right vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. If there exists a point $P$ on the ellipse, different from $A_{1}$ and $A_{2}$, such that $\overrightarrow{P O} \cdot \overrightarrow{P A_{2}}=0$, where $O$ is the origin, then the range of the eccentricity $e$ of the ellipse is ( ).
(A) $\left(0, \frac{1}{2}\right)$
(B) $\left(0, \frac{\sqrt{2}}{2}\right)$
(C) $\left(\frac{1}{2}, 1\right)$
(D) $\left(\frac{\sqrt{2}}{2}, 1\right)$
7. D.

From the given, we know $\angle O P A_{2}=90^{\circ}$.
Let $P(x, y)(x>0)$.
The equation of the circle with $\mathrm{OA}_{2}$ as its diameter is
$$
\left(x-\frac{a}{2}\right)^{2}+y^{2}=\frac{a^{2}}{4} \text {, }
$$

Combining this with the ellipse equation, we get
$$
\left(1-\frac{b^{2}}{a^{2}}\right) x^{2}-a x+b^{2}=0 \text {. }
$$

From the given, we know that this equation has real roots in $(0, a)$.
This leads to $0\frac{1}{2} \text {. }
$$

Therefore, the range of $e$ is $\left(\frac{\sqrt{2}}{2}, 1\right)$.