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Current Question (ID: 16478)

Question:
$\text{In a hydrogen atom, the electron and proton are bound at a distance of about } 0.53 \, \text{\AA}. \text{ The potential energy of the system in eV is:}$ $\text{(taking the zero of the potential energy at an infinite separation of the electron from the proton.)}$
Options:
  • 1. $-23.1 \, \text{eV}$
  • 2. $27.0 \, \text{eV}$
  • 3. $-27.2 \, \text{eV}$
  • 4. $23.7 \, \text{eV}$
Solution:
$\text{Hint: } U = -\frac{k q_1 q_2}{r}$ $\text{Step: Find the electrostatic potential energy of the system.}$ $\text{The electrostatic potential energy between two charges } q_1 \text{ and } q_2 \text{ separated by a distance } r \text{ is given by:}$ $U = -\frac{k q_1 q_2}{r}$ $\text{Substitute the values we get,}$ $U = -\frac{(9 \times 10^9) \times (1.602 \times 10^{-19})^2}{0.53 \times 10^{-10}} = -43.5 \times 10^{-19} \, \text{J}$ $\text{Now, convert the result to electron volts:}$ $U = \frac{-43.5 \times 10^{-19}}{1.6 \times 10^{-19}} = -27.2 \, \text{eV}$ $\text{Hence, option (3) is the correct answer.}$

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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
}