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Current Question (ID: 20905)

Question:

$\text{The ratio of the shortest wavelengths in the Lyman and Balmer series of the H-atom is:}$

Options:

1. $\frac{1}{4}$
2. $\frac{4}{1}$
3. $\frac{1}{2}$
4. $\frac{2}{1}$

Solution:

$\text{The Lyman and Balmer series are spectral series in the hydrogen atom, representing electron transitions:}$ $\text{the Lyman series involves transitions to the } n=1 \text{ energy level (ultraviolet region),}$ $\text{while the Balmer series involves transitions to } n=2 \text{ (visible light region).}$ $\text{The shortest wavelength in any spectral series occurs when the electron transitions}$ $\text{from the highest energy level } (n_2 \rightarrow \infty) \text{ to the lowest energy level of the series.}$ $\text{For the Lyman series (transitions to } n_1=1):$ $\frac{1}{(\lambda_{\text{shortest}})_{\text{Lyman}}} = R(1 - 0)$ $\text{For the Balmer series (transitions to } n_1=2):$ $\frac{1}{(\lambda_{\text{shortest}})_{\text{Balmer}}} = R\left(\frac{1}{4} - 0\right)$ $\text{So,}$ $(\lambda_{\text{shortest}})_{\text{Lyman}} = \frac{1}{R}; \ (\lambda_{\text{shortest}})_{\text{Balmer}} = \frac{4}{R}$ $\frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}} = \frac{\frac{1}{R}}{\frac{4}{R}} = \frac{1}{4}$ $\text{Hence, option 1 is the correct answer.}$

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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
}