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Current Question (ID: 7485)

Question:
$\text{A radiation source emitted } 2.5 \times 10^{15} \text{ photons in 2 nanoseconds. Energy of the source is :}$
Options:
  • 1. $8.28 \times 10^{-10} \text{ J}$
  • 2. $8.24 \times 10^{10} \text{ J}$
  • 3. $7.12 \times 10^{-11} \text{ J}$
  • 4. $2.12 \times 10^{10} \text{ J}$
Solution:
$\text{Hint: Frequency} = \frac{1}{\text{T}}$\n\n$\text{Step 1:}$\n\n$\text{Frequency of radiation } (\nu),$\n\n$\nu = \frac{1}{2.0} \times 10^{-9} \text{ s} = 5.0 \times 10^8 \text{ s}^{-1}$\n\n$\text{Step 2:}$\n\n$\text{Energy (E) of source} = \text{Nh}\nu$\n\n$\text{Where,}$\n\n$\text{N = number of photons emitted}$\n\n$\text{h = Planck's constant, } \nu \text{ = frequency of radiation}$\n\n$\text{Substituting the values in the given expression of (E):}$\n\n$\text{E} = (2.5 \times 10^{15})(6.626 \times 10^{-34} \text{ Js})(5 \times 10^8 \text{ s}^{-1})$\n\n$= 8.28 \times 10^{-10} \text{ J}$\n\n$\text{Hence, the energy of the source (E) is } 8.28 \times 10^{-10} \text{ J}$

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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
}