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Current Question (ID: 19244)

Question:

$\text{If the potential energy between two molecules is given by}$ $U = -\frac{A}{r^6} + \frac{B}{r^{12}},$ $\text{then the potential energy at equilibrium separation between molecules is:}$

Options:

1. $-\frac{A^2}{2B}$
2. $-\frac{A^2}{4B}$
3. $0$
4. $-\frac{A^2}{3B}$

Solution:

$\text{Hint: At equilibrium position, } F_{\text{net}} = 0$ $\text{Step 1: Find the force between the molecules}$ $F = -\frac{dU}{dr}$ $= -\left[ \frac{6A}{r^7} - \frac{12B}{r^{13}} \right]$ $\text{Step 2: Find the equilibrium separation.}$ $\frac{6A}{r^7} - \frac{12B}{r^{13}} = 0$ $\Rightarrow r^6 = \frac{12B}{6A}$ $r^6 = \frac{2B}{A}$ $\text{Step 3: Find the potential energy.}$ $U = -\frac{A \times A}{2B} + \frac{B \times A^2}{4B^2} = -\frac{A^2}{4B}$

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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
}