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Current Question (ID: 20884)

Question:

$\text{Given that an electron in an atom moves at } 600 \text{ m/s with an accuracy of } 0.005\%, \text{ what is the level of certainty in locating its position?}$ $\text{(h = } 6.6 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}, \text{ mass of electron, } e_m = 9.1 \times 10^{-31} \text{ kg)}:$

Options:

1. $1.52 \times 10^{-4} \text{ m}$
2. $5.10 \times 10^{-3} \text{ m}$
3. $1.92 \times 10^{-3} \text{ m}$
4. $3.84 \times 10^{-3} \text{ m}$

Solution:

$\text{Hint: The formula is } \Delta x \geq \frac{h}{4\pi m \Delta v}$ $\text{Step 1:}$ $\text{The given values are as follows:}$ $\text{Speed of electron } = 600 \text{ m/s}$ $\text{Accuracy of speed } = 0.005\%$ $\text{Uncertainty in speed } (\Delta V) = 600 \times \frac{0.005}{100} = 0.03 \text{ m/s}$ $\text{Step 2:}$ $\text{According to Heisenberg's uncertainty principle:}$ $\Delta x \Delta p \geq \frac{h}{4\pi}$ $\Delta x \times m \Delta v \geq \frac{h}{4\pi}$ $\Delta x \geq \frac{h}{4\pi m \Delta v}$ $\Delta x \geq \frac{6.6 \times 10^{-34} \text{ Js}}{4 \times 3.14 \times 9.1 \times 10^{-31} \text{ kg} \times 0.03 \text{ m/s}}$ $\Delta x \geq 1.9248 \times 10^{-3} \text{ m}$ $\text{Thus, option third is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}