5. Given that vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ are two unit vectors in a plane with an angle of $60^{\circ}$ between them, and $(2 \boldsymbol{\alpha}-\boldsymbol{\gamma}) \cdot(\boldsymbol{\beta}-\boldsymbol{\gamma})=0$, then the maximum value of $|\gamma|$ is $\qquad$ .
5. $\frac{\sqrt{7}+\sqrt{3}}{2}$.

Let $\overrightarrow{O A}=2 \boldsymbol{\alpha}, \overrightarrow{O B}=2 \boldsymbol{\beta}, \overrightarrow{O C}=2 \boldsymbol{\gamma}$. From the given information, we have $A C \perp B C$. Therefore, point $C$ lies on the circle $\odot O_{1}$ with $A B$ as its diameter.
When $O C$ passes through $O_{1}$, $O C$ is maximized. By the cosine rule, we know
$$
A B=\sqrt{2^{2}+1^{2}-2 \times 2 \times 1 \times \cos 60^{\circ}}=\sqrt{3} .
$$

Thus, $\angle A B O=90^{\circ}$. Hence,
$$
O O_{1}=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}+1}=\frac{\sqrt{7}}{2},
$$

Therefore, the maximum value of $O C$ is
$$
O O_{1}+O_{1} C=\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}=\frac{\sqrt{7}+\sqrt{3}}{2} .
$$