Import Question JSON

Current Question (ID: 16785)

Question:

$\text{A neutron, a proton, an electron and an } \alpha\text{-particle enter a region of the uniform magnetic field with the same velocity.}$ $\text{The magnetic field is perpendicular and directed into the plane of the paper.}$ $\text{The tracks of the particles are labelled in the figure.}$ $\text{Which track will the } \alpha\text{-particle follow?}$

Options:

1. $A$
2. $B$
3. $C$
4. $D$

Solution:

$\text{Hint: } \textbf{R} = \frac{mv}{Bq}$ $\text{Step: Find the path of the different particles.}$ $\text{If the velocity } v \text{ is perpendicular to the magnetic field } B, \text{ the magnetic force is perpendicular to both } v \text{ and } B \text{ and acts like a centripetal force.}$ $\text{When the particle is moving in a uniform magnetic field then the radius of the circle described by the charged particle is given by:}$ $r = \frac{mv}{qB}$ $\text{When the velocity of the particle and magnetic field of the particle are constant then;}$ $r \propto \frac{m}{q}$ $\text{For neutron, } r_n \propto \frac{m}{0} = \infty \text{ means neutron remains undeflected in a magnetic field.}$ $\text{For electron, } r_e \propto \frac{m_e}{e} \approx \text{ very small value}$ $\text{For proton, } r_p \propto \frac{4m_e}{e}$ $\text{For } \alpha\text{-particle, } r_\alpha \propto \frac{2 \times (4m_e)}{2e} \propto 2r_p$ $\text{Therefore, in the plot A corresponds to electron, B is for the neutron, C for the } \alpha\text{-particle, and D for the proton}$ $\text{Hence, option (3) is the correct answer.}$

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
}