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Current Question (ID: 7491)

Question:

The work function for the Cesium atom is 1.9 eV. The threshold frequency of the radiation is:

Options:

1. $4.59 \times 10^{14} \, \text{s}^{-1}$
2. $8.59 \times 10^{14} \, \text{s}^{-1}$
3. $5.59 \times 10^{-14} \, \text{s}^{-1}$
4. $65.9 \times 10^{14} \, \text{s}^{-1}$

Solution:

### **Hint:** Photoelectric effect. **Explanation:** **Step 1:** Calculation of threshold frequency. From the expression, $W_0 = h\nu_0$, we get: \[ \nu_0 = \frac{W_0}{h} \] Where $\nu_0$ = threshold frequency, $h$ = Planck’s constant **Step 2:** Substituting the values in the given expression of $\nu_0$: \[ \nu_0 = \frac{1.602 \times 10^{-19} \times 1.9}{6.626 \times 10^{-34}} \] (1 eV = $1.602 \times 10^{-19} \, \text{J}$) \[ \nu_0 = 4.59 \times 10^{14} \, \text{s}^{-1} \] Hence, the threshold frequency of radiation ($\nu_0$) is $4.59 \times 10^{14} \, \text{s}^{-1}$.

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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
}