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Current Question (ID: 20892)

Question:

$\text{According to Bohr's atomic theory:}$ $\text{(A) Kinetic energy of the electron is } \propto \frac{Z^2}{n^2}$ $\text{(B) The product of velocity (v) of electron and principal quantum number (n), } vn' \propto Z^2.$ $\text{(C) Frequency of revolution of the electron in an orbit is } \propto \frac{Z^3}{n^3}.$ $\text{(D) Coulombic force of attraction on the electron is } \propto \frac{Z^3}{n^4}.$ $\text{Choose the most appropriate answer from the options given below:}$

Options:

1. $(\text{C}) \text{ Only}$
2. $(\text{A}) \text{ Only}$
3. $(\text{A}), (\text{C}) \text{ and } (\text{D}) \text{ only}$
4. $(\text{A}) \text{ and } (\text{D}) \text{ only}$

Solution:

$\text{Hint: frequency } \propto \frac{z^2}{n^3}$ $\text{According to Bohr's theory}$ $1. \text{ KE } = 13.6 \frac{Z^2}{n^2} \frac{eV}{\text{atom}} \Rightarrow \text{ KE } \propto \frac{Z^2}{n^2}$ $2. \text{ speed of } e^- \propto \frac{z}{n}$ $\cdot v \times n \propto z$ $3. \text{ Frequency of revolution of } e^- = \frac{v}{2\pi r}$ $\cdot \text{ frequency } \propto \frac{z^3}{n^3}$ $4. \text{ F } = \frac{k q_1 q_2}{r^2} = \frac{kze^2}{r^2} \left\{ r \alpha \frac{n^2}{z} \right\}$ $\Rightarrow \text{ F } \alpha \left( \frac{z}{n^2} \right)^2$ $\Rightarrow \text{ F } \alpha \frac{z^3}{n^4}$

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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
}