Import Question JSON

Current Question (ID: 7515)

Question:

$\text{Assertion (A): The angular momentum of an electron in an atom is quantized.}$\n\n$\text{Reason (R): In an atom, only those orbits are permitted in which the angular momentum of the electron is a natural number multiple of } h/2\pi.$

Options:

1. $\text{Both (A) and (R) are True and (R) is the correct explanation of (A).}$
2. $\text{Both (A) and (R) are True but (R) is not the correct explanation of (A).}$
3. $\text{(A) is True but (R) is False.}$
4. $\text{(A) is False but (R) is True.}$

Solution:

$\text{Hint: The angular momentum of an electron is quantised.}$\n\n$\text{In an atom, an electron can move only in those orbits for which its angular momentum is an integral multiple of } h/2\pi. \text{ That means angular momentum is quantised. Radiation is emitted or absorbed only when transition of electrons takes place from one quantised value of angular momentum to another.}$\n\n$\text{According to Bohr's model, the angular momentum of an electron in an orbit is given by:}$\n\n$mvr = n\frac{h}{2\pi}$\n\n$\text{where } m \text{ is the mass of the electron, } v \text{ is its velocity, } r \text{ is the radius of the orbit, } n \text{ is a positive integer (quantum number), and } h \text{ is Planck's constant.}$\n\n$\text{This equation shows that the angular momentum of an electron in an atom is quantized to discrete values, with each value being an integral multiple of } h/2\pi.$\n\n$\text{Therefore, the assertion (A) is true, and the reason (R) is also true. Furthermore, (R) directly explains why (A) is true - the angular momentum is quantized precisely because only orbits with angular momentum equal to an integral multiple of } h/2\pi \text{ are permitted.}$\n\n$\text{Hence, both (A) and (R) are true, and (R) is the correct explanation of (A).}$

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
}