Import Question JSON

Current Question (ID: 7513)

Question:

$\text{A dye absorbs a photon of wavelength } \lambda \text{ and re-emits the same energy into two photons of wavelengths } \lambda_1 \text{ and } \lambda_2 \text{ respectively. The wavelength } \lambda \text{ is related to } \lambda_1 \text{ and } \lambda_2 \text{ as:}$

Options:

1. $\lambda = \frac{\lambda_1+\lambda_2}{\lambda_1\lambda_2}$
2. $\lambda = \frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2}$
3. $\lambda = \frac{\lambda_1^2+\lambda_2^2}{\lambda_1+\lambda_2}$
4. $\lambda = \frac{\lambda_1\lambda_2}{(\lambda_1+\lambda_2)^2}$

Solution:

$\text{Hint: Total energy = sum of the energy of two photons}$\n\n$\text{Step 1:}$\n\n$\text{The total energy of a photon is E and wavelength is } \lambda. \text{ The energy of two photons are } E_1 \text{ and } E_2 \text{ and wavelength is } \lambda_1 \text{ and } \lambda_2.$\n\n$\text{The formula of E is as follows:}$\n\n$E = \frac{hc}{\lambda}$\n\n$\text{Step 2:}$\n\n$\text{The total energy (E) = } E_1 + E_2$\n\n$\frac{hc}{\lambda} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$\n\n$\frac{hc}{\lambda} = hc \left( \frac{\lambda_2+\lambda_1}{\lambda_1\lambda_2} \right)$\n\n$\text{Dividing both sides by } hc:$\n\n$\frac{1}{\lambda} = \frac{\lambda_1 + \lambda_2}{\lambda_1\lambda_2}$\n\n$\lambda = \frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2}$

Import JSON File

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
}