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Current Question (ID: 10779)

Question:

\text{The wavelength corresponding to maximum emission of radiation from the sun is } \lambda_{\max} = 4753 \text{ Å (close to the wavelength of violet colour of visible region).} \text{Hence if temperature is doubled, } \lambda_m \text{ is decreased } \left(\lambda_m \propto \frac{1}{T}\right) \text{ i.e. mostly ultraviolet radiations emits.} \text{If the temperature of the sun becomes twice its present temperature, then:}

Options:

1. $\text{Radiated energy would be predominantly in the infrared range.}$
2. $\text{Radiated energy would be primarily in the ultraviolet range.}$ (Correct)
3. $\text{Radiated energy would be predominantly in the X-ray region}$
4. $\text{Radiated energy would become twice as strong as it is now.}$

Solution:

\text{This question is based on Wien's displacement law, which states that } \lambda_{\text{max}} \propto \frac{1}{T} \text{Given information:} \text{Current } \lambda_{\text{max}} = 4753 \text{ Å (violet region of visible spectrum)} \text{Temperature is doubled: } T_{\text{new}} = 2T_{\text{original}} \text{According to Wien's law:} \lambda_{\text{max,new}} = \frac{\lambda_{\text{max,original}}}{2} = \frac{4753}{2} = 2376.5 \text{ Å} \text{This wavelength (2376.5 Å) falls in the ultraviolet range of the electromagnetic spectrum.} \text{Therefore, when the temperature is doubled, the peak wavelength shifts from the violet region to the ultraviolet region, meaning the radiated energy would be primarily in the ultraviolet range.}

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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
}