Import Question JSON

Current Question (ID: 7523)

Question:

$\text{If uncertainty in position and momentum are both equal, then uncertainty in velocity will be:}$

Options:

1. $\frac{1}{2m} \sqrt{\frac{h}{\pi}}$
2. $\sqrt{\frac{h}{2\pi}}$
3. $\frac{1}{m} \sqrt{\frac{h}{\pi}}$
4. $\frac{h}{\pi}$

Solution:

$\text{Hint: } \Delta x \times \Delta p \geq \frac{h}{4\pi}$\n\n$\text{Explanation:}$\n\n$\text{According to Heisenberg's uncertainty principle, it is impossible to determine simultaneously the position and momentum of a moving microscopic particle.}$\n\n$\text{The formula is as follows:}$\n\n$\Delta x \times \Delta p \geq \frac{h}{4\pi}$\n\n$\text{Given that } \Delta x = \Delta p \text{, and we know that } \Delta p = m \cdot \Delta v \text{, where } m \text{ is the mass of the particle.}$\n\n$\text{Substituting these into the uncertainty principle:}$\n\n$\Delta x \times m \cdot \Delta v \geq \frac{h}{4\pi}$\n\n$\text{Since } \Delta x = \Delta p = m \cdot \Delta v\text{:}$\n\n$m \cdot \Delta v \times m \cdot \Delta v \geq \frac{h}{4\pi}$\n\n$m^2 \cdot (\Delta v)^2 \geq \frac{h}{4\pi}$\n\n$(\Delta v)^2 \geq \frac{h}{4\pi m^2}$\n\n$\Delta v \geq \frac{1}{2m} \sqrt{\frac{h}{\pi}}$\n\n$\text{Therefore, the uncertainty in velocity will be } \frac{1}{2m} \sqrt{\frac{h}{\pi}}$

Import JSON File

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
}