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Current Question (ID: 7529)

Question:
$\text{The graphs that represents the variation of momentum of particle with de-Broglie wavelength is:}$
Options:
  • 1. $\text{Graph showing p vs } \lambda \text{ as a straight line with negative slope}$
  • 2. $\text{Graph showing p vs } \lambda \text{ as a curve with increasing slope}$
  • 3. $\text{Graph showing } \lambda \text{ vs } \frac{1}{p} \text{ as a curve with decreasing slope}$
  • 4. $\text{Graph showing } \lambda \text{ vs } \frac{1}{p} \text{ as a straight line with positive slope}$
Solution:
$\text{Hint: } \lambda = \frac{h}{\text{mv}}$ $\text{From de Broglie's hypothesis, the momentum of a particle, p = h/}\lambda$ $\text{where h = Plank's constant and } \lambda \text{ = wavelength.}$ $\text{The relation between wavelength and } \frac{1}{\text{p}} \text{ is inverse.}$ $\text{So, the graph between the de Broglie wavelength and } \frac{1}{\text{p}} \text{ of a photon is a straight line because it follows a straight line equation, that is, y=mx.}$ $\text{According to de Broglie's equation: } p = \frac{h}{\lambda}$ $\text{This can be rewritten as: } \lambda = \frac{h}{p}$ $\text{This shows an inverse relationship between momentum (p) and wavelength (}\lambda\text{).}$ $\text{When p is plotted on y-axis and }\lambda\text{ on x-axis (as in option 1), we get a hyperbola (negative slope curve).}$ $\text{Since } \lambda \propto \frac{1}{p}\text{, the correct graph would show p decreasing as }\lambda\text{ increases.}$ $\text{Option 1 correctly shows this inverse relationship with a curve having negative slope.}$

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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
}