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Current Question (ID: 17954)

Question:

$\text{If 'a' stands for the edge length of the cubic systems: simple cubic, body-centered cubic, and face-centered cubic, then the ratio of radii of the spheres in these systems will be respectively:}$

Options:

1. $\frac{1}{2}a : \frac{\sqrt{3}}{4}a : \frac{1}{2\sqrt{2}}a$
2. $\frac{1}{2}a : \sqrt{3}a : \frac{1}{\sqrt{2}}a$
3. $\frac{1}{2}a : \frac{\sqrt{3}}{2}a : \frac{\sqrt{2}}{2}a$
4. $1a : \sqrt{3}a : \sqrt{2}a$

Solution:

$\text{HINT: For FCC, } a = 2\sqrt{2}r \text{ or } r = \frac{a}{2\sqrt{2}}$ $\text{Explanation:}$ $\text{For simple cubic, we have}$ $a = 2r \text{ or } r = \frac{a}{2}$ $\text{For body centred cubic, we have}$ $a = \frac{4r}{\sqrt{3}} \text{ or } r = \frac{\sqrt{3}}{4}a$ $\text{For face centred cubic, we have}$ $a = 2\sqrt{2}r \text{ or } r = \frac{a}{2\sqrt{2}}$ $\text{Hence, the ratio of radii in simple cubic, body centred cubic and face centred cubic is:}$ $\frac{a}{2} : \frac{\sqrt{3}}{4}a : \frac{a}{2\sqrt{2}}$

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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
}