1. Let $\alpha, \beta$ be a pair of conjugate complex numbers. If $|\alpha-\beta|=2 \sqrt{3}$ and $\frac{\alpha}{\beta^{2}}$ is a real number, then $|\alpha|=$ $\qquad$ .
1. Let $\alpha=a+b \mathrm{i}, \beta=a-b \mathrm{i}, a, b \in \mathbb{R}$, then from $|\alpha-\beta|=2 \sqrt{3}$, we get $|b|=\sqrt{3}$.
From $\frac{\alpha}{\beta^{2}}=\frac{\alpha^{3}}{(\alpha \beta)^{2}}$ being a real number and $\alpha \beta$ being a real number, we know that $\alpha^{3}$ is a real number.
That is, $(a+b \mathrm{i})^{3}=\left(a^{3}-3 a b^{2}\right)+\left(3 a^{2} b-b^{3}\right) \mathrm{i}$ is a real number,
thus $3 a^{2} b-b^{3}=0$, hence we get $|a|=1$,
$$
\therefore \quad|\alpha|=\sqrt{a^{2}+b^{2}}=2 .
$$