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Current Question (ID: 7504)

Question:

The maximum wavelength in the Lyman series of $\text{He}^{+}$ ion is-

Options:

1. $3\text{R}$
2. $1/(3\text{R})$
3. $1/\text{R}$
4. $2\text{R}$

Solution:

Hint: Rydberg equation-based calculation of wavelength in hydrogen spectrum Here, use the Rydberg equation for the calculation of maximum wavelength in the Lyman series of $\text{He}^{+}$ ion. The Rydberg equation is given by: $\frac{1}{\lambda} = \text{R} \text{Z}^2 \left( \frac{1}{\text{n}_1^2} - \frac{1}{\text{n}_2^2} \right)$ For the Lyman series, the electron transitions to the ground state, so $\text{n}_1 = 1$. To find the maximum wavelength ($\lambda_{\text{max}}$), the energy difference must be the smallest. This occurs for the transition from the next higher energy level to $\text{n}_1$, which means $\text{n}_2 = 2$. For $\text{He}^{+}$ ion, the atomic number $\text{Z} = 2$. Substituting these values into the Rydberg equation: $\frac{1}{\lambda_{\text{max}}} = \text{R} (2)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$ $\frac{1}{\lambda_{\text{max}}} = \text{R} \times 4 \left( \frac{1}{1} - \frac{1}{4} \right)$ $\frac{1}{\lambda_{\text{max}}} = 4\text{R} \left( \frac{4-1}{4} \right)$ $\frac{1}{\lambda_{\text{max}}} = 4\text{R} \left( \frac{3}{4} \right)$ $\frac{1}{\lambda_{\text{max}}} = 3\text{R}$ Therefore, $\lambda_{\text{max}} = \frac{1}{3\text{R}}$ The final answer is $\boxed{2}$

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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
}