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Current Question (ID: 7495)

Question:

The ratio of slopes of $K_{\text{max}}$ vs. $v$ and $V_0$ vs. $v$ curves in photoelectric effect gives - ($v$=frequency, $K_{\text{max}}= \text{maximum kinetic energy}$, $V_0$=stopping potential):

Options:

1. Charge of electron
2. Planck's constant
3. Work function
4. The ratio of Planck's constant to electronic charge

Solution:

Hint: Apply the concept of Photoelectric effect Step 1: For the stopping potential (V_0) vs frequency (v) curve: eV_0 = hv - hv_0 V_0 = (h/e)v - (h/e)v_0 The slope (slope_1) is h/e. Step 2: For the maximum kinetic energy (K_max) vs frequency (v) curve: K_max = hv - hv_0 The slope (slope_2) is h. Step 3: The ratio of slopes: slope_2/slope_1 = h/(h/e) = e where e is the electronic charge. Thus, the ratio gives the charge of the electron.

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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
}