Import Question JSON

Current Question (ID: 7584)

Question:

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

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:

$\text{Hint: Energy bound to the nucleus is the difference between the energy of}$ $\text{the incident photon and the kinetic energy of the electron.}$\n\n$\text{Explanation:}$\n\n$\text{Step 1:}$\n\n$\text{Calculate the energy of the incident photon}$\n\n$\text{The energy of the incident photon (E) is given by,}$\n\n$E = \frac{hc}{\lambda}$\n\n$= \frac{(6.626) \times 10^{-34} \times (3.0) \times 10^8 \text{ ms}^{-1} \times (150) \times 10^{-12} \text{ m}}{1}$\n\n$= 1.3252 \times 10^{-15} \text{ J}$\n\n$= 13.252 \times 10^{-16} \text{ J}$\n\n$\text{Step 2:}$\n\n$\text{Calculate the energy of the electron ejected}$\n\n$\text{The energy of the electron ejected (K.E) = } \frac{1}{2}mv^2$\n\n$= \frac{1}{2}(9.10939) \times 10^{-31} \text{ kg } (1.5) \times 10^7 \text{ ms}^{-1})^2$\n\n$= 10.24 \times 10^{-17} \text{ J}$\n\n$= 1.02 \times 10^{-16} \text{ J}$\n\n$\text{Step 3:}$\n\n$\text{Find out the energy bound to the nucleus.}$\n\n$\text{Hence, the energy with which the electron is bound to the nucleus can be}$ $\text{obtained as:}$\n\n$= E - KE$\n\n$= 13.252 \times 10^{-16} \text{ J} - 1.025 \times 10^{-16} \text{ J}$\n\n$= 12.22 \times 10^{-16} \text{ J}$\n\n$\text{Hence, option second is the correct answer.}$

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
}