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Current Question (ID: 7487)

Question:

Electromagnetic radiation of wavelength $242$ nm is just sufficient to ionise sodium atom. The ionisation energy of sodium in kJ mol$^{-1}$ is :

Options:

1. $494$
2. $4.94$
3. $516$
4. $0.50$

Solution:

### Hint: Energy of sodium $(E) = \frac{N_A hc}{\lambda}$ ## STEP 1: Write down the given data Wavelength $(\lambda) = 242 \, \text{nm} = 242 \times 10^{-9}$ m ## STEP 2: Calculate Ionisation energy of sodium Ionisation Energy of sodium $(E) = \frac{N_A hc}{\lambda}$ $= 6.023 \times 10^{23} \, \text{mol}^{-1} \times 6.626 \times 10^{-34} \, \text{Js} \times 3 \times 10^8 \, \text{ms}^{-1} \, 242 \times 10^{-9}$ m $= 494.7 \times 10^3 \, \text{Jmol}^{-1}$ $= 494 \, \text{kJ mol}^{-1}$

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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
}