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Current Question (ID: 7502)

Question:

The transition in the hydrogen spectrum that would have the same wavelength as Balmer transition from $n = 4$ to $n = 2$ of $\text{He}^+$ spectrum is :

Options:

1. $n_1 = 3$ to $n_2 = 4$
2. $n_2 = 3$ to $n_1 = 2$
3. $n_2 = 3$ to $n_1 = 1$
4. $n_2 = 2$ to $n_1 = 1$

Solution:

**Hint:** \[(\overline{v}) = \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\] **Step 1:** Find out the wavelength for Balmer Series. For $\text{He}^+$ ion, the wave number $(\overline{v})$ associated with the Balmer transition, $n = 4$ to $n = 2$ is given by: \[(\overline{v}) = \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\] Where, $n_1 = 2$, $n_2 = 4$ $Z = 2$ (atomic number of helium) \[(\overline{v}) = \frac{1}{\lambda} = R(2)^2 \left( \frac{1}{4} - \frac{1}{16} \right)\] \[\overline{v} = \frac{1}{\lambda} = \frac{3R}{4}\] \[\Rightarrow \lambda = \frac{4}{3R}\] **Step 2:** Find the equivalent transition in hydrogen spectrum. For hydrogen ($Z=1$), we need to find a transition with the same wavelength: \[\frac{1}{\lambda} = R(1^2) \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3R}{4}\] \[\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4}\] By testing the options, this equality holds true only when $n_1 = 1$ and $n_2 = 2$. ∴ The transition for $n_2 = 2$ to $n_1 = 1$ in hydrogen spectrum would have the same wavelength as Balmer transition $n = 4$ to $n = 2$ of $\text{He}^+$ spectrum.

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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
}