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Current Question (ID: 16479)

Question:

$\text{An elementary particle of mass } m \text{ and charge } +e \text{ is projected with velocity } v$ $\text{at a much more massive particle of charge } Ze, \text{ where } Z > 0.$ $\text{What is the closest possible approach of the incident particle?}$

Options:

1. $\frac{Ze^2}{2\pi\varepsilon_0 mv^2}$
2. $\frac{Ze}{4\pi\varepsilon_0 mv^2}$
3. $\frac{Ze^2}{8\pi\varepsilon_0 mv^2}$
4. $\frac{Ze}{8\pi\varepsilon_0 mv^2}$

Solution:

$\text{Suppose distance of closest approach is } r, \text{ and according to energy}$ $\text{conservation applied for elementary charge.}$ $\text{Energy at the time of projection = Energy at the distance of closest approach}$ $\Rightarrow \frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(Ze) \cdot e}{r} \Rightarrow r = \frac{Ze^2}{2\pi\varepsilon_0 mv^2}$

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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
}