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Current Question (ID: 7542)

Question:

$\text{The excited state of an H atom emits a photon of wavelength } \lambda \text{ and returns to the ground state. The principal quantum number of the excited state is given by:}$

Options:

1. $\sqrt{\lambda R(\lambda R - 1)}$
2. $\sqrt{\frac{\lambda R}{(\lambda R - 1)}}$
3. $\sqrt{\lambda R(\lambda R + 1)}$
4. $\sqrt{\frac{(\lambda R - 1)}{\lambda R}}$

Solution:

$\text{Hint: } \frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ $\text{Using the Rydberg formula for hydrogen atom:}$ $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ $\text{where } n_1 = 1 \text{ (ground state), } n_2 = \text{? (excited state)}$ $\text{Substituting } n_1 = 1:$ $\frac{1}{\lambda} = R\left(\frac{1}{1} - \frac{1}{n_2^2}\right) = R\left(1 - \frac{1}{n_2^2}\right)$ $\text{Rearranging to solve for } n_2^2:$ $\frac{1}{\lambda} = R - \frac{R}{n_2^2}$ $\frac{R}{n_2^2} = R - \frac{1}{\lambda}$ $\frac{R}{n_2^2} = \frac{R\lambda - 1}{\lambda}$ $n_2^2 = \frac{R\lambda}{R\lambda - 1} = \frac{\lambda R}{\lambda R - 1}$ $\text{Therefore: } n_2 = \sqrt{\frac{\lambda R}{\lambda R - 1}}$ $\text{The correct answer is option 2.}$

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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
}