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Current Question (ID: 7516)

Question:
$\text{For any H-like system, the ratio of velocities of I, II & III orbit i.e., } V_1 : V_2 : V_3 \text{ will be:}$
Options:
  • 1. $1 : 2 : 3$
  • 2. $1 : 1/2 : 1/3$
  • 3. $3 : 2 : 1$
  • 4. $1 : 1 : 1$
Solution:
$\text{Hint: Apply Bohr's model}$\n\n$\text{According to Bohr's model, the velocity of an electron in the } n^{th} \text{ orbit of a hydrogen-like system is given by:}$\n\n$v = 2.18 \times 10^6 \frac{Z}{n} \text{ m/s}$\n\n$\text{where } Z \text{ is the atomic number and } n \text{ is the principal quantum number.}$\n\n$\text{From this equation, we can see that } v \propto \frac{1}{n}$\n\n$\text{For the first orbit (} n = 1 \text{):}$\n\n$v_1 \propto \frac{1}{1} = 1$\n\n$\text{For the second orbit (} n = 2 \text{):}$\n\n$v_2 \propto \frac{1}{2}$\n\n$\text{For the third orbit (} n = 3 \text{):}$\n\n$v_3 \propto \frac{1}{3}$\n\n$\text{Therefore, the ratio of velocities is:}$\n\n$v_1 : v_2 : v_3 = 1 : \frac{1}{2} : \frac{1}{3}$

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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
}