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Current Question (ID: 7532)

Question:

$\text{The kinetic energy of an electron is } 3.0 \times 10^{-25} \text{ J. Its wave length would be:}$

Options:

1. $8.96 \times 10^{-7} \text{ m}$
2. $4.37 \times 10^{-6} \text{ m}$
3. $1.32 \times 10^{-7} \text{ m}$
4. $2.89 \times 10^{-4} \text{ m}$

Solution:

$\text{HINT: Use de-Broglie Equation.}$ $\text{STEP 1: Write down de-Broglie Equation}$ $\lambda = \frac{h}{mv}$ $\text{where, } \lambda = \text{wavelength of light; h = Planck's constant; m = mass of particle and v = velocity of light}$ $\text{STEP 2: Find out velocity from the Kinetic energy}$ $\text{K.E.} = \frac{1}{2}mv^2$ $\text{Given, K.E.} = 3.0 \times 10^{-25} \text{ J}$ $\text{So, v} = \left(\frac{2 \times 3.0 \times 10^{-25}}{9.11 \times 10^{-31}}\right)^{1/2}$ $\text{On solving we get, v} = 811.57 \text{ m/s}$ $\text{STEP 3: Find wavelength using de-Broglie Equation}$ $\lambda = \frac{6.63 \times 10^{-34} \text{ Js}}{9.11 \times 10^{-31} \text{ kg} \times 811.57 \text{ ms}^{-1}} = 8.96 \times 10^{-7} \text{ m}$ $\text{Hence, the wavelength of the electron is } 8.96 \times 10^{-7} \text{ m}$

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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
}