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Current Question (ID: 7507)

Question:

$\text{The wave number for the longest wavelength transition in the Balmer series of atomic hydrogen would be:}$

Options:

1. $1.52 \times 10^6 m^{-1}$
2. $3.14 \times 10^6 (cm)^{-1}$
3. $15.2 \times 10^6 m^{-1}$
4. $1.52 \times 10^6 (cm)^{-1}$

Solution:

$\text{HINT: Use Rydberg's formula}$\n\n$\text{STEP 1: Write down the general formula to find out wavenumber of Balmer series.}$\n\n$\text{For the Balmer series, } n_i = 2. \text{ Thus, the expression of wavenumber } (\bar{\nu}) \text{ is given by,}$\n\n$\bar{\nu} = \left[\frac{1}{2^2} - \frac{1}{n_f^2}\right]\left(1.09 \times 10^7 m^{-1}\right)$\n\n$\text{STEP 2: Find out } n_f \text{ value for the given condition and calculate wavenumber.}$\n\n$\text{Wave number } (\bar{\nu}) \text{ is inversely proportional to wavelength of transition.}$\n\n$\text{Hence, for the longest wavelength transition, } (\bar{\nu}) \text{ has to be the smallest.}$\n\n$\text{For } (\bar{\nu}) \text{ to be minimum, } n_f \text{ should be minimum.}$\n\n$\text{For the Balmer series, a transition from } n_i = 2 \text{ to } n_f = 3 \text{ is allowed.}$\n\n$\text{Hence, taking } n_f = 3, \text{ we get:}$\n\n$\bar{\nu} = \left(1.09 \times 10^7 \text{ m}^{-1}\right)\left[\frac{1}{2^2} - \frac{1}{3^2}\right]$\n\n$= \left(1.09 \times 10^7 \text{ m}^{-1}\right)\left[\frac{1}{4} - \frac{1}{9}\right]$\n\n$= 1.52 \times 10^6 \text{ m}^{-1}$

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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
}