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Current Question (ID: 7483)

Question:
$\text{A 25-watt bulb emits monochromatic yellow light with a wave length of 0.57μm. The rate of emission of quanta per second would be:}$
Options:
  • 1. $7.17 \times 10^{-19} \text{ s}^{-1}$
  • 2. $4.13 \times 10^{16} \text{ s}^{-1}$
  • 3. $7.17 \times 10^{19} \text{ s}^{-1}$
  • 4. $1.26 \times 10^{20} \text{ s}^{-1}$
Solution:
$\text{Hint: Rate of emission of quanta per second} = \frac{\text{Power}}{\text{Energy}}$ $\text{STEP 1: Find out the energy using given wavelength.}$ $\text{Power of bulb, P = 25 Watt = 25 J/s}$ $\text{Energy of one photon, E = h}\nu = \frac{hc}{\lambda}$ $\text{Substituting the values in the given expression of E:}$ $E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.57 \times 10^{-6}} = 34.87 \times 10^{-20} \text{ J}$ $\text{STEP 2: Find out rate of emission of quanta per second.}$ $\text{Rate of emission of quanta per second}$ $= \frac{25 \text{ J s}^{-1}}{34.87 \times 10^{-20} \text{ J}} = 7.17 \times 10^{19} \text{ s}^{-1}$

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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
}