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Current Question (ID: 7524)

Question:

$\text{If the position of the electron were measured with an accuracy of } \pm 0.002 \text{ nm, the uncertainty in the momentum of the electron would be:}$

Options:

1. $5.637 \times 10^{-23} \text{ kg m s}^{-1}$
2. $4.637 \times 10^{-23} \text{ kg m s}^{-1}$
3. $2.637 \times 10^{-23} \text{ kg m s}^{-1}$
4. $3.637 \times 10^{-23} \text{ kg m s}^{-1}$

Solution:

$\text{Hint: Use Heisenberg's uncertainty principle.}$\n\n$\text{Explanation:}$\n\n$\text{Step 1: Write down the expression of Heisenberg's uncertainty principle.}$\n\n$\text{From Heisenberg's uncertainty principle:}$\n\n$\Delta x \times \Delta p = \frac{h}{4\pi} \Rightarrow \Delta p = \frac{1}{\Delta x} \cdot \frac{h}{4\pi} \text{ ................(I)}$\n\n$\text{Where,}$\n\n$\Delta x = \text{uncertainty in the position of the electron}$\n\n$\Delta p = \text{uncertainty in the momentum of the electron}$\n\n$\text{Step 2: Substitute the values in equation (I)}$\n\n$\text{Given: } \Delta x = 0.002 \text{ nm} = 0.002 \times 10^{-9} \text{ m}$\n\n$\text{Planck's constant, } h = 6.626 \times 10^{-34} \text{ J s}$\n\n$\Delta p = \frac{6.626 \times 10^{-34}}{0.002 \text{ nm} \times 4 \times 3.14}$\n\n$\Delta p = \frac{6.626 \times 10^{-34}}{0.002 \times 10^{-9} \times 4 \times 3.14}$\n\n$\Delta p = \frac{6.626 \times 10^{-34}}{2.512 \times 10^{-11}}$\n\n$\Delta p = 2.637 \times 10^{-23} \text{ J s m}^{-1}$\n\n$\text{Since } 1 \text{ J} = 1 \text{ kg m}^2 \text{ s}^{-2}\text{:}$\n\n$\Delta p = 2.637 \times 10^{-23} \text{ kg m s}^{-1}$\n\n$\text{Therefore, the uncertainty in the momentum of the electron is } 2.637 \times 10^{-23} \text{ kg m s}^{-1}$

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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
}