In an atom, an electron is moving with a speed of $$600$$ m/s with an accuracy of $$0.005\%$$. Certainity with which the position of the electron can be located is ($$h = 6.6 \times 10^{-34}$$ kg m$$^2$$ s$$^{-1}$$, mass of electron, $$e_m = 9.1 \times 10^{-31}$$ kg)

For simultaneous measurement of position and momentum, Heisenberg’s uncertainty principle states
$$\Delta x\,\Delta p \;\ge\; \dfrac{h}{4\pi}$$

The electron’s momentum uncertainty is related to its speed uncertainty by $$\Delta p = m_e\,\Delta v$$.

Step 1: Determine the speed uncertainty.
Given speed $$v = 600\,$$m s$$^{-1}$$ with an accuracy of $$0.005\% = 0.005/100 = 5.0\times10^{-5}$$, therefore
$$\Delta v = v \times 5.0\times10^{-5} = 600 \times 5.0\times10^{-5} = 0.03\,$$m s$$^{-1}$$.

Step 2: Calculate the momentum uncertainty.
Electron mass $$m_e = 9.1\times10^{-31}\,$$kg, hence
$$\Delta p = m_e \,\Delta v = 9.1\times10^{-31}\times0.03 = 2.73\times10^{-32}\,\text{kg m s}^{-1}.$$

Step 3: Apply the uncertainty relation to find positional uncertainty.
$$\Delta x \;\ge\; \dfrac{h}{4\pi\,\Delta p} = \dfrac{6.6\times10^{-34}}{4\pi \times 2.73\times10^{-32}}.$$

Compute the denominator first:
$$4\pi \times 2.73\times10^{-32} \approx 12.566 \times 2.73\times10^{-32} = 3.43\times10^{-31}.$$

Hence
$$\Delta x \;\ge\; \dfrac{6.6\times10^{-34}}{3.43\times10^{-31}} \approx 1.92\times10^{-3}\,\text{m}.$$

The minimum uncertainty in the electron’s position is therefore $$1.92 \times 10^{-3}\,$$m.

Option C which is: $$1.92 \times 10^{-3}\,$$m