The minimum uncertainty in the speed of an electron in one dimensional region of length $$2a_0$$ (Where $$a_0$$ = Bohr radius = 52.9 pm) is _____ km s$$^{-1}$$ (Nearest integer) (Given: Mass of electron $$= 9.1 \times 10^{-31}$$ kg, Planck's constant $$h = 6.63 \times 10^{-34}$$ Js)

We need to find the minimum uncertainty in the speed of an electron confined to a one-dimensional region of length $$2a_0$$, where $$a_0 = 52.9$$ pm is the Bohr radius.

We use the Heisenberg Uncertainty Principle in one dimension:

$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$

Here, the uncertainty in position $$\Delta x = 2a_0$$. For the minimum uncertainty, we take the equality:

$$\Delta p = \frac{h}{4\pi \cdot \Delta x} = \frac{h}{4\pi \cdot 2a_0}$$

Since $$\Delta p = m \cdot \Delta v$$, we have:

$$\Delta v = \frac{h}{4\pi \cdot 2a_0 \cdot m}$$

Now we substitute the values. We have $$h = 6.63 \times 10^{-34}$$ J s, $$a_0 = 52.9$$ pm $$= 52.9 \times 10^{-12}$$ m, and $$m = 9.1 \times 10^{-31}$$ kg.

$$\Delta v = \frac{6.63 \times 10^{-34}}{4\pi \times 2 \times 52.9 \times 10^{-12} \times 9.1 \times 10^{-31}}$$

Let us compute the denominator step by step. We have $$4\pi \approx 12.566$$, and:

$$12.566 \times 2 \times 52.9 \times 10^{-12} \times 9.1 \times 10^{-31}$$

$$= 12.566 \times 105.8 \times 10^{-12} \times 9.1 \times 10^{-31}$$

$$= 12.566 \times 962.78 \times 10^{-43}$$

$$= 12101.4 \times 10^{-43}$$

$$= 1.21014 \times 10^{-39}$$

Now dividing:

$$\Delta v = \frac{6.63 \times 10^{-34}}{1.21014 \times 10^{-39}} = \frac{6.63}{1.21014} \times 10^{5}$$

$$= 5.479 \times 10^{5} \text{ m s}^{-1}$$

Converting to km/s: $$\Delta v = 547.9 \approx 548$$ km s$$^{-1}$$.

Hence, the correct answer is 548.