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Current Question (ID: 7482)

Question:

$\text{The wavelength of the light emitted when the electron returns to the ground state in the H atom, from n = 5 to n = 1, would be: (The ground-state electron energy is } -2.18 \times 10^{-18} \text{ j)}$

Options:

1. $\text{Wavelength} = 9.498 \times 10^{-8} \text{ km}$
2. $\text{Wavelength} = 12.498 \times 10^{-8} \text{ m}$
3. $\text{Wavelength} = 9.498 \times 10^{-8} \text{ m}$
4. $\text{Wavelength} = 9.498 \times 10^{-8} \text{ cm}$

Solution:

$\text{HINT: 1st calculate energy and then wavelength can be calculated.}$ $\text{STEP 1: Write down the given data and general expression of Energy}$ $\text{Energy (E) of the n}^{\text{th}} \text{ Bohr orbit of an atom is given by,}$ $E_n = \frac{-(2.18)}{1} \times 10^{-18} Z^2/n^2$ $\text{Where,}$ $Z = \text{atomic number of the atom}$ $\text{Ground state energy} = -2.18 \times 10^{-11} \text{ ergs}$ $= -2.18 \times 10^{-11} \times 10^{-7} \text{ J}$ $= -2.18 \times 10^{-18} \text{ J}$ $\text{STEP 2: Find out } \Delta E$ $\text{Energy required to shift the electron from n = 1 to n = 5 is given as:}$ $\Delta E = E_5 - E_1$ $= \frac{-(2.18)}{1} \times 10^{-18} \left( \times (1^2)(1/5)^2 \right) - (-2.18) \times 10^{-18}$ $= (2.18) \times 10^{-18} \left[1 - \frac{1}{25}\right]$ $= (2.18) \times 10^{-18} \left( \frac{24}{25} \right) = 2.0928 \times 10^{-18} \text{ J}$ $\text{STEP 3: Calculate } \lambda \text{ by using } \Delta E$ $\text{Wavelength of emitted light} = \frac{hc}{E}$ $= \frac{(6.626) \times 10^{-34}(3) \times 10^8}{(2.0928) \times 10^{-18}}$ $= 9.498 \times 10^{-8} \text{ m}$

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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
}