Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter $$10^{-15} \text{ m}$$ is ______ $$\times 10^9 \text{ ms}^{-1}$$ (nearest integer) [Given : mass of electron $$= 9.1 \times 10^{-31} \text{ kg}$$, Planck's constant $$(h) = 6.626 \times 10^{-34} \text{ Js}$$] (Value of $$\pi = 3.14$$)

By Heisenberg's uncertainty principle: $$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$.

$$\Delta v = \frac{h}{4\pi m \Delta x} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15}}$$

$$= \frac{6.626 \times 10^{-34}}{114.296 \times 10^{-46}} = \frac{6.626}{114.296} \times 10^{12} = 0.05798 \times 10^{12} = 5.798 \times 10^{10}$$

$$= 57.98 \times 10^9 \approx 58 \times 10^9$$ m/s.

The answer is $$\boxed{58}$$.