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Current Question (ID: 7499)

Question:

When an electron jumps from n=5 to n=1 in a hydrogen atom, the number of spectral lines obtained is

Options:

1. 3
2. 4 (Correct)
3. 6
4. 10

Solution:

### **Hint:** Use the formula for number of possible transitions between energy levels **Step 1:** The number of possible spectral lines when an electron falls from energy level n to level m is given by the combination formula: \[ \text{Number of transitions} = \frac{(n - m)(n - m + 1)}{2} \] **Step 2:** For transition from n=5 to n=1: \[ \text{Number of transitions} = \frac{(5 - 1)(5 - 1 + 1)}{2} = \frac{4 \times 5}{2} = 10 \] **Alternative Explanation:** The electron can make direct or indirect transitions through intermediate levels: - 4 transitions to n=1 (5→1, 4→1, 3→1, 2→1) - 3 transitions to n=2 (5→2, 4→2, 3→2) - 2 transitions to n=3 (5→3, 4→3) - 1 transition to n=4 (5→4) Total = 4 + 3 + 2 + 1 = 10 transitions Thus, the correct answer is 10 spectral lines.

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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
}