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Current Question (ID: 17989)

Question:

$\text{Higher conductivity of p-type and n-type semiconductors is due to:}$

Options:

1. $\text{Increase in the number of positive holes and mobile electrons respectively.}$
2. $\text{Increase in the number of electrons in both types of semiconductors.}$
3. $\text{Increase in the number of positive-holes in both types of semiconductors.}$
4. $\text{None of the above.}$

Solution:

$\text{HINT: n-type semiconductors have negatively charged electrons as charge carriers.}$ $\text{Explanation:}$ $\text{The two main types of semiconductors are:}$ $\text{(i) n-type semiconductor (ii) p-type semiconductor}$ $\text{n-type semiconductor: The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor.}$ $\text{When the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As, an n-type semiconductor is generated.}$ $\text{p-type semiconductor: The semiconductor whose increased in conductivity is a result of electron-hole is called a p-type semiconductor.}$ $\text{When a crystal of group 14 elements such as Si or Ge is doped with a group 13 element such as B, Al, or Ga (which contains only three valence electrons), a p-type of semiconductor is generated.}$

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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
}