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Current Question (ID: 7510)

Question:

$\text{The ionization energy for H atom in the ground state is } 2.18 \times 10^{-18} \text{ J. The process energy requirements will be: } (He^{+}(g) \rightarrow He^{2+}(g) + e^{-})$

Options:

1. $8.72 \times 10^{-18} \text{ J}$
2. $7.54 \times 10^{-18} \text{ J}$
3. $5.67 \times 10^{-17} \text{ J}$
4. $2.18 \times 10^{-17} \text{ J}$

Solution:

$\text{Hint: } \text{E}_n = -2.18 \times 10^{-18} \left(\frac{Z^2}{n^2}\right) \text{ J}$\n\n$\text{Step 1:}$\n\n$\text{The energy formula for hydrogen or hydrogen-like species is as follows:}$\n\n$\text{E}_n = -2.18 \times 10^{-18} \left(\frac{Z^2}{n^2}\right) \text{ J}$\n\n$\text{For the ground state of the hydrogen atom, } \Delta\text{E} = \text{E}_{\infty} - \text{E}_1 = 2.18 \times 10^{-18} \text{ J}$\n\n$\text{Step 2:}$\n\n$\text{For the given process, } \text{He}^{+}(\text{g}) \rightarrow \text{He}^{2+}(\text{g}) + \text{e}^{-}$\n\n$\text{An electron is removed from n = 1 to n = } \infty, \text{ Z = 2 for He}^{+}$\n\n$\Delta E = E_{\infty} - E_1$\n\n$= 0 - \left[-2.18 \times 10^{-18} \left(\frac{2^2}{1^2}\right) \text{ J}\right]$\n\n$= 0 - \left[-2.18 \times 10^{-18} \times 4 \text{ J}\right]$\n\n$= 0 - \left[-8.72 \times 10^{-18} \text{ J}\right]$\n\n$= 8.72 \times 10^{-18} \text{ J}$

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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
}