Import Question JSON

Current Question (ID: 7492)

Question:

When a photon with a wavelength of $150$ pm strikes an atom, one of its inner bound electrons is ejected at a velocity of $1.5 \times 10^7 \, \text{m} \, \text{s}^{-1}$. The energy with which it is bound to the nucleus would be:

Options:

1. $32.22 \times 10^{-16} \, \text{J}$
2. $12.22 \times 10^{-16} \, \text{J}$
3. $22.27 \times 10^{-16} \, \text{J}$
4. $31.22 \times 10^{-16} \, \text{J}$

Solution:

Step 1: Calculate the energy of the incident photon using the formula: E_photon = hc / lambda where h = 6.626 * 10^(-34) Js, c = 3 * 10^8 m/s, and lambda = 150 * 10^(-12) m. Step 2: Calculate the kinetic energy of the ejected electron using: KE = (1/2) * m_e * v^2 where m_e = 9.109 * 10^(-31) kg and v = 1.5 * 10^7 m/s. Step 3: The binding energy is the difference between the photon energy and the electron's kinetic energy: E_binding = E_photon - KE Step 4: Substituting the values: E_photon = (6.626 * 10^(-34)) * (3 * 10^8) / (150 * 10^(-12)) = 1.3252 * 10^(-15) J KE = (1/2) * (9.109 * 10^(-31)) * (1.5 * 10^7)^2 = 1.023 * 10^(-16) J E_binding = 1.3252 * 10^(-15) - 1.023 * 10^(-16) = 1.2229 * 10^(-15) J = 12.229 * 10^(-16) J Thus, the correct answer is 12.22 * 10^(-16) J.

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
}