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Current Question (ID: 16780)

Question:
$\text{An electron is moving in a circular path under the influence of a transverse magnetic field of } 3.57 \times 10^{-2} \text{ T.}$ $\text{If the value of } \frac{e}{m} \text{ is } 1.76 \times 10^{11} \text{ C/kg,}$ $\text{what will be the frequency of revolution of the electron?}$
Options:
  • 1. $1 \text{ GHz}$
  • 2. $100 \text{ MHz}$
  • 3. $62.8 \text{ MHz}$
  • 4. $6.28 \text{ MHz}$
Solution:
$\text{Hint: The magnetic force provides the required centripetal force.}$ $\text{Step 1: Find the radius of the circular path.}$ $\text{The radius of the revolving electron is given by,}$ $r = \frac{mv}{qB}$ $\text{Step 2: Find the time period of revolution.}$ $\text{Time table for a complete cycle,}$ $T = \frac{2\pi}{v} = \frac{2\pi}{v} \times \frac{mv}{qB} = \frac{2\pi m}{qB}$ $\text{Step 3: Find the frequency of revolution.}$ $\text{Frequency, } f = \frac{1}{T} = \frac{eB}{2\pi m} = \frac{1.76 \times 10^{11} \times 3.57 \times 10^{-2}}{2\pi} = 1 \text{ GHz}$

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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
}