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Current Question (ID: 7496)

Question:

Determine the maximum number of emission lines produced when an electron in a hydrogen atom transitions from the $n = 6$ energy level to the ground state:

Options:

1. 30
2. 21
3. 15
4. 28

Solution:

### **HINT:** Use the general formula to calculate number of spectral lines **STEP 1:** The number of possible emission lines when an electron transitions from energy level $n$ to the ground state is given by: \[\text{Number of spectral lines} = \frac{n(n-1)}{2}\] **STEP 2:** For $n = 6$: \[\text{Number of spectral lines} = \frac{6(6-1)}{2} = \frac{6 \times 5}{2} = 15\] Thus, the maximum number of emission lines is 15.

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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
}