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Current Question (ID: 7509)

Question:

$\text{The transition in the He}^{+} \text{ ion in the balmer series that would have the same wave number as the first Lyman line in the hydrogen spectrum is-}$

Options:

1. $2 \rightarrow 1$
2. $5 \rightarrow 3$
3. $4 \rightarrow 2$
4. $6 \rightarrow 4$

Solution:

$\text{Hint: Both transitions have an equal wavenumber.}$\n\n$\text{The formula which relates wavelength and rydberg constant is}$\n\n$\frac{1}{\lambda} = RZ^2 \left[ \frac{1}{(n_1)^2} - \frac{1}{(n_2)^2} \right]$\n\n$\text{For the first Lyman series, the } n_1 \text{ value is 1 and } n_2 \text{ value is 2. The equation of wavelength is}$\n\n$\text{For } 1^{st} \text{ Lyman series of H} = \frac{1}{\lambda} = RZ^2 \left[ \frac{1}{(1)^2} - \frac{1}{(2)^2} \right]$\n\n$\text{For Balmer series, } n_1 = 2, \text{ and } n_2 \text{ we have to find}$\n\n$\text{For transition of He}^{+} \text{ in Balmer series for He}^{+} = RZ^2 \frac{1}{(2)^2} - \frac{1}{(n_2)^2}$\n\n$\text{Equate the two equations as follows:}$\n\n$\text{For H}$\n\n$\frac{1}{\lambda_1} = R \cdot (1)^2 \cdot \left[ \frac{1}{(1)^2} - \frac{1}{(2)^2} \right]$\n\n$= \frac{3R}{4}$\n\n$\text{For He}^{+}$\n\n$\frac{1}{\lambda_2} = R \cdot (2)^2 \cdot \frac{1}{(2)^2} - \frac{1}{n_2^2}$\n\n$\frac{1}{\lambda_2} = 4R \times \left[ \frac{1}{(2)^2} - \frac{1}{n_2^2} \right]$\n\n$\text{Since } \lambda_1 = \lambda_2$\n\n$\frac{3R}{4} = 4R - \left[ \frac{1}{4} - \frac{1}{n_2^2} \right]$\n\n$\frac{1}{n_2^2} = \frac{1}{4} - \frac{3}{16}$\n\n$\frac{1}{n_2^2} = \frac{1}{16}$\n\n$n_2 = 4$

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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
}