Question

Uncertainty in the Position of an Electron (mass $9.1\times {10}^{-31}\mathrm{kg}$) moving with a velocity of $300{\mathrm{ms}}^{-1}$, Accurate up to $0.001\%$ will be:

Answer 1

Step 1: Heisenberg’s Uncertainty Principle:

• The uncertainty principle is the consequence of the dual behavior of matter and radiation.
• It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.
• Mathematically, it can be given as in the equation:
• $∆x\times ∆p_x\ge \frac{h}{4\pi }$, where $∆x$ is the uncertainty in position and $∆p_x(\mathrm{or}∆v_x)$ is the uncertainty in momentum (or velocity) of the particle.

Step 2: Given Information:

• $\mathrm{m}$= $9.1\times {10}^{-31}\mathrm{kg}$
• $\mathrm{v}$= $300{\mathrm{ms}}^{-1}$

Step 3: Calculation for uncertainty in velocity:

$$\begin{gathered} ∆{\mathrm{v}}_{\mathrm{x}}=\frac{300\times 0.001}{100} \\ ∆{\mathrm{v}}_{\mathrm{x}}=3\times {10}^{-3}{\mathrm{ms}}^{-1} \end{gathered}$$

Step 4: Calculation for uncertainty in position:

$$\begin{gathered} ∆\mathrm{x}=\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{u}} \\ ∆\mathrm{x}=\frac{6.626\times {10}^{-34}\times 100}{4\times 3.14\times 9.1\times {10}^{-31}\times 0.001\times 300} \\ ∆\mathrm{x}=1.93\mathrm{x}{10}^{-2}\mathrm{m} \end{gathered}$$

Thus, the uncertainty in the position of electron is $1.93\times {10}^{-2}\mathrm{m}$.