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Current Question (ID: 7505)

Question:

Emission transitions in the Paschen series end at orbit $\text{n} = 3$ and start from orbit $\text{n}$ and can be represented as $\nu = 3.29 \times 10^{15} \left( \frac{1}{3^2} - \frac{1}{\text{n}^2} \right) \text{ Hz}$. The value of $\text{n}$ if the transition is observed at $1285 \text{ nm}$ is:

Options:

1. 6
2. 5
3. 8
4. 9

Solution:

Hint: Use the given formula of frequency Explanation: Step 1: Calculate the frequency from wavelength and (c) velocity of light Given, $\lambda = 1285 \text{ nm} = 1285 \times 10^{-9} \text{ m}$ We know that the relationship between frequency ($\nu$), wavelength ($\lambda$), and the speed of light ($\text{c}$) is: $\nu = \frac{\text{c}}{\lambda}$ Using $\text{c} = 3 \times 10^8 \text{ m s}^{-1}$: $\nu = \frac{3 \times 10^8 \text{ m s}^{-1}}{1285 \times 10^{-9} \text{ m}}$ $\nu \approx 2.33 \times 10^{14} \text{ s}^{-1}$ or $\text{Hz}$ Step 2: Substitute the value of frequency in the given equation and calculate $\text{n}$. Given, $\nu = 3.29 \times 10^{15} \left( \frac{1}{3^2} - \frac{1}{\text{n}^2} \right)$ Thus, $2.33 \times 10^{14} = 3.29 \times 10^{15} \left( \frac{1}{3^2} - \frac{1}{\text{n}^2} \right)$ Divide both sides by $3.29 \times 10^{15}$: $\frac{2.33 \times 10^{14}}{3.29 \times 10^{15}} = \frac{1}{3^2} - \frac{1}{\text{n}^2}$ $0.0708 \approx \frac{1}{9} - \frac{1}{\text{n}^2}$ $0.0708 \approx 0.1111 - \frac{1}{\text{n}^2}$ Rearrange to solve for $\frac{1}{\text{n}^2}$: $\frac{1}{\text{n}^2} \approx 0.1111 - 0.0708$ $\frac{1}{\text{n}^2} \approx 0.0403$ Take the reciprocal of both sides: $\text{n}^2 \approx \frac{1}{0.0403}$ $\text{n}^2 \approx 24.81$ Take the square root: $\text{n} \approx \sqrt{24.81}$ $\text{n} \approx 4.98 \approx 5$ Hence, for the transition to be observed at $1285 \text{ nm}$, $\text{n} = 5$. The spectrum lies in the infra-red region. The final answer is $\boxed{2}$

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{
  "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
}