Import Question JSON

Current Question (ID: 7530)

Question:

$\text{Incorrect statement among the following is:}$

Options:

1. $\text{The uncertainty principle is } \Delta x \cdot \Delta p \geq \frac{h}{4\pi}$
2. $\text{Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry, and a more balanced arrangement.}$
3. $\text{The energy of the 2s orbital is less than the energy of the 2p orbital in the case of hydrogen-like atoms.}$
4. $\text{De-Broglie's wavelength is given by } \lambda = \frac{h}{mv}\text{, where m = mass of the particle, v = group velocity of the particle.}$

Solution:

$\text{Hint: Apply Heisenberg's Uncertainty principle}$ $\text{Explanation:}$ $\text{In the case of the hydrogen atom, the energies of the 2s-orbital and 2p-orbital are equal. This can be explained as follows:}$ $\text{Energy for hydrogen-like species can be given as,}$ $\text{E} = -13.6 \frac{\text{Z}^2}{\text{n}^2}\text{ eV}$ $\text{For 2s and 2p, n=2, so energy will be the same for 2s and 2p orbitals.}$ $\text{All other are correct statements. Thus, option 3 is the correct answer.}$ $\text{Option 1 is correct: The Heisenberg's uncertainty principle states that the product of uncertainties in position and momentum is equal to or greater than h/4π.}$ $\text{Option 2 is correct: Half-filled and fully filled orbitals indeed have greater stability due to the mentioned factors.}$ $\text{Option 4 is correct: De Broglie's wavelength formula is correctly stated with the proper relationship between wavelength, Planck's constant, mass, and velocity.}$

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
}