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Current Question (ID: 16406)

Question:

$\text{Suppose the charge of a proton and an electron differ slightly. One of them is } -e, \text{ the other is } (e + \Delta e).$ $\text{If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance } d \text{ (much greater than atomic size) apart is zero, then } \Delta e \text{ is of the order of?}$ $\text{(Given the mass of hydrogen } m_h = 1.67 \times 10^{-27} \text{ kg)}$

Options:

1. $10^{-23} \text{ C}$
2. $10^{-37} \text{ C}$
3. $10^{-47} \text{ C}$
4. $10^{-20} \text{ C}$

Solution:

$\text{Hint: } F_G + F_r = 0$ $\text{Step: Write the equation for electrostatic force and gravitational force.}$ $\text{The net charge on one H-atom is given by;} \ q = -e + e + \Delta e = \Delta e$ $\text{The net electrostatic repulsive force between two H-atoms is given by;} \ F_r = \frac{kq^2}{d^2} = \frac{k\Delta e^2}{d^2}$ $\text{Similarly, the net gravitational attraction force between the two H-atoms is given by;} \ F_G = \frac{Gm^2}{d^2}$ $\text{According to the question, the net force on the atom is zero we have,}$ $\frac{Gm^2}{d^2} - \frac{k\Delta e^2}{d^2} = 0$ $\Delta e^2 = \frac{Gm^2}{k}$ $\Delta e^2 = \frac{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2}{9 \times 10^9}$ $\Rightarrow \Delta e = 1.437 \times 10^{-37} \text{ C}$ $\text{Hence, option (2) is the correct answer.}$

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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
}