$$\sqrt{d_1}$$ and $$\sqrt{d_2}$$ are the impact parameters corresponding to scattering angles 60° and 90° respectively, when an $$\alpha$$ particle is approaching a gold nucleus. For $$d_1 = xd_2$$, the value of $$x$$ will be ______.

In Rutherford scattering, the impact parameter $$b$$ is related to the scattering angle $$\theta$$ by the relation $$b = \frac{a}{2}\cot\Bigl(\frac{\theta}{2}\Bigr)$$, where $$a$$ denotes the distance of closest approach. When the scattering angle is $$\theta_1 = 60°$$, the corresponding impact parameter satisfies $$\sqrt{d_1} = \frac{a}{2}\cot(30°) = \frac{a}{2}\times\sqrt{3}$$ and for $$\theta_2 = 90°$$ it becomes $$\sqrt{d_2} = \frac{a}{2}\cot(45°) = \frac{a}{2}\times1$$.

The ratio of these two impact parameters can be written as $$\frac{\sqrt{d_1}}{\sqrt{d_2}} = \frac{\frac{a}{2}\sqrt{3}}{\frac{a}{2}} = \sqrt{3}$$ and, upon squaring both sides, one obtains $$\frac{d_1}{d_2} = 3$$. Hence, it follows that $$d_1 = 3\,d_2$$, which directly leads to the result $$x = 3$$.

Therefore, the value of $$x$$ is 3.