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Current Question (ID: 16776)

Question:
$\text{Moving perpendicular to field } B, \text{ a proton and an alpha particle both enter an area of uniform magnetic field } B. \text{ If the kinetic energy of the proton is } 1 \text{ MeV and the radius of the circular orbits for both particles is equal, the energy of the alpha particle will be:}$
Options:
  • 1. $4 \text{ MeV}$
  • 2. $0.5 \text{ MeV}$
  • 3. $1.5 \text{ MeV}$
  • 4. $1 \text{ MeV}$
Solution:
$\text{Hint: The alpha particle has two times the charge of the proton and four times the mass of the proton.}$ $\text{Step 1: Find the formula of energy.}$ $\text{Radius in the circular path is given by,}$ $R = \frac{mv}{qB} = \frac{\sqrt{2mE}}{qB}$ $\text{And, the total energy of a moving particle in a circular orbit,}$ $E = \frac{q^2B^2R^2}{2m}$ $\text{Step 2: Find the energy of the proton.}$ $\text{For a proton entering a region of the magnetic field,}$ $E_1 = \frac{e^2B^2R^2}{2m_p} \quad \text{....(1)}$ $\text{where } m_p \text{ is the mass of proton.}$ $\text{Step 3: Find the energy of the alpha particle.}$ $\text{Similarly, for an } \alpha\text{-particle moving in a uniform magnetic field,}$ $E_2 = \frac{(2e)^2B^2R^2}{2(4m_p)} \quad \left[\because m_\alpha = 4m_p\right] \quad \text{...(2)}$ $\text{Dividing Eq. (ii) by eq. (i) we get,}$ $E_2 = E_1 = 1 \text{ MeV}$

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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
}