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Current Question (ID: 20878)

Question:

$\text{The radius of the second Bohr orbit for hydrogen atom is:}$ $\text{(Planck's Const. } h = 6.6262 \times 10^{-34} \text{ Js; mass of electron } = 9.1091 \times 10^{-31} \text{ kg;}$ $\text{charge of electron } e = 1.60210 \times 10^{-19} \text{ C; permittivity of vacuum } \varepsilon_0 = 8.854185 \times 10^{-12} \text{ kg}^{-1}\text{m}^{-3}\text{A}^2)}$

Options:

1. $0.529 \text{ \AA}$
2. $2.12 \text{ \AA}$
3. $1.65 \text{ \AA}$
4. $7.76 \text{ \AA}$

Solution:

$\text{Hint: Use the formula of radius, that is, } = 0.529 \times \frac{n^2}{Z} \text{ \AA}$ $\text{The formula of radius is as follows:}$ $r = 0.529 \times \frac{n^2}{Z} \text{ \AA}$ $= 0.529 \times \frac{(2)^2}{1}$ $= 2.12 \text{ \AA}$

Import JSON File

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
}