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Current Question (ID: 16793)

Question:

$\text{In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential } V \text{ and then made to describe semi-circular paths of radius } R \text{ using a magnetic field } B. \text{ If } V \text{ and } B \text{ are kept constant, the ratio of } \left( \frac{\text{Charge on the ion}}{\text{Mass of the ion}} \right) \text{ will be proportional to:}$

Options:

1. $\frac{1}{R}$
2. $\frac{1}{R^2}$
3. $R^2$
4. $R$

Solution:

$\text{Hint: The centripetal force is provided by the magnetic force.}$ $\text{Step 1: Find the magnetic force on the ions.}$ $\text{The magnetic force} = qvB = \frac{mv^2}{R}$ $\text{Step 2: Find the energy gained by the ions.}$ $\text{The angular frequency is given by,}$ $\omega = \frac{v}{R} = \frac{qB}{m}$ $\text{Energy of ion is given by,}$ $E = \frac{1}{2} mv^2 = \frac{1}{2} m (R \omega)^2 = \frac{1}{2} m R^2 B^2 \frac{q^2}{m^2}$ $\text{Or } E = \frac{1}{2} \frac{R^2 B^2 q^2}{m} \quad \text{(i)}$ $\text{If ions are accelerated by electric potential } V, \text{ then energy attained by ions,}$ $E = qV \quad \text{(ii)}$ $\text{Step 3: Equate both of them.}$ $\text{From Eqs. (i) and (ii), we get,}$ $qV = \frac{1}{2} \frac{R^2 B^2 q^2}{m}$ $\text{or } \frac{q}{m} = \frac{1}{2V} \frac{R^2 B^2}{R^2 B^2}$ $\text{If } V \text{ and } B \text{ are kept constant, then,}$ $\frac{q}{m} \propto \frac{1}{R^2}$

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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
}