Import Question JSON

« Back to Question List Edit Question »

Current Question (ID: 7920)

Question:
$\text{The energy of an electron in the ground state of the hydrogen atom is } -2.18 \times 10^{-18} \text{ J. The ionization enthalpy of atomic hydrogen in terms of J mol}^{-1} \text{ is:}$
Options:
  • 1. $2.81 \times 10^6 \text{ J mol}^{-1}$
  • 2. $1.31 \times 10^6 \text{ J mol}^{-1}$ (Correct)
  • 3. $2.31 \times 10^6 \text{ J mol}^{-1}$
  • 4. $1.81 \times 10^6 \text{ J mol}^{-1}$
Solution:
$\text{HINT: Apply the idea of the mole concept to derive the answer.}$\n\n$\text{Explanation:}$\n\n$\text{Ionization enthalpy is a measure of the amount of energy required to remove an electron from an atom or ion in its gaseous state. In the hydrogen atom, the electron is present in the 1s orbital.}$\n\n$\text{The energy of an electron in the ground state of the hydrogen atom is equal to the ionization enthalpy of the hydrogen atom but opposite in sign.}$\n\n$\text{The energy of an electron in the ground state of the hydrogen atom is } -2.18 \times 10^{-18} \text{ J. Therefore, the energy required to remove that electron from the ground state of a hydrogen atom is } 2.18 \times 10^{-18} \text{ J.}$\n\n$\text{Ionization enthalpy of atomic hydrogen = } 2.18 \times 10^{-18} \text{ J. Hence, ionization enthalpy of atomic hydrogen in terms of J mol}^{-1} \text{ is}$\n\n$= 2.18 \times 10^{-18} \times 6.02 \times 10^{23} \text{ J mol}^{-1}$\n\n$= 1.31 \times 10^6 \text{ J mol}^{-1}$\n\n$\text{Key steps:}$\n\n$\text{1. Ground state energy of electron = } -2.18 \times 10^{-18} \text{ J}$\n\n$\text{2. Ionization energy per atom = } +2.18 \times 10^{-18} \text{ J (opposite sign)}$\n\n$\text{3. Convert to per mole by multiplying with Avogadro's number}$\n\n$\text{4. Final answer = } 1.31 \times 10^6 \text{ J mol}^{-1}$

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
}